'     /vi.  W/r  /"/'<n)  /^M 
t~Th S.I /■'<■<>  '"^- 


ANALYTIC    GEOMETRY 


BY 


W.   A.   WILSON,   Ph.D. 

ASSISTANT    PROFESSOR   OF   MATHEMATICS 
YALE   UNIVERSITY 

AND 

J.   I.   TRACEY,   Ph.D. 

INSTRUCTOR   IN   MATHEMATICS 
YALE   UNIVERSITY 


3i*;< 


D.    C.    HEATH    &   CO.,   PUBLISHERS 
BOSTON  NEW  YORK  CHICAGO 


■ . ' '    '  / 

Copyright,  191 5, 
By  D.  C.   Heath  &  Co. 

2E  2 


■Engineering. 


PREFACE 

The  purpose  of  this  book  is  to  present  in  a  short  course 
those  parts  of  Analytic  Geometry  which  are  essential  for 
the  study  of  Calculus.  The  material  has  been  so  arranged 
that  topics  which  are  less  important  may  be  omitted  with- 
out a  loss  of  continuity,  and  the  text  is  therefore  adapted 
for  use  in  classes  which  aim  to  cover  in  one  year  the  funda- 
mental principles  and  applications  of  both  Analytic  Geom- 
etry and  Calculus.  For  such  classes  Chapters  I-VI,  VIII, 
and  the  earlier  sections  of  Chapters  VII  and  IX  form  a 
course  which  can  be  completed  in  about  thirty-five  lessons. 
In  order  to  provide  material  for  study  if  more  time  is  avail- 
able, a  discussion  of  the  general  equation  of  the  second 
degree  and  short  chapters  on  Tangents  and  Normals  and 
Solid  Analytic  Geometry  have  been  included. 

In  the  preparation  of  the  book  the  authors  have  used 
as  a  foundation  a  pamphlet  written  by  Professor  William 
Beebe  for  use  in  his  classes.  The  authors  wish  to  express 
their  thanks  to  Dr.  David  D.  Leib  and  to  Dr.  Levi  L.  Conant 
for  many  valuable  criticisms  and  suggestions. 


6S4544 


m 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/analyticgeometryOOwilsrich 


CONTENTS 


Introduction 


CHAPTER   I  — COORDINATES   AND 
EQUATIONS 

SECTION 

1.  Introduction 7 

2.  Position ;  Cartesian  Coordinates 7 

3.  Notation 8 

4.  Plotting  of  Points 9 

5.  Directed  Lines 11 

6.  The  Distance  Formula 12 

7.  The  Point  of  Division  Formulas 13 

8.  Geometric  Applications 14 

9.  Angle  between  Two  Lines 16 

10.  Inclination  and  Slope 17 

11.  The  Slope  Formula 17 

12.  Parallelism  and  Perpendicularity 18 

13.  The  Angle  Formula 18 

14.  Equations  and  Graphs 20 

15.  Plotting  of  Graphs 21 

16.  Derivation  of  Equations 24 

17.  Functional  Variables 27 

18.  Functional  Notation 28 

19.  Graphs  of  Functions 29 

20.  Discussion  of  Equations 30 

21.  Intercepts 31 

22.  Symmetry 31 

23.  Extent _       .        .  33 

24.  Symmetrical  Transformations 35 

25.  Transformations  of  Equations  not  Altering  their  Loci    .  36 

26.  Equations  whose  Graphs  cannot  be  Plotted      ...  37 

27.  Intersection  of  Curves 38 

V 


VI 


CONTENTS 


CHAPTER   II  — THE   STEAIGHT   LINE 

SECTION 

28.  Equations  of  the  Straight  Line 

29.  The  Point  Slope  and  Two  Point  Forms 

30.  The  Slope  Intercept  Form 

31.  The  Intercept  Form  .... 

32.  Lines  Parallel  to  the  Axes 

33.  The  Linear  Equation 

34.  Relations  between  Linear  Equations 

35.  Application  to  the  Solution  of  Linear  Equations 

36.  Geometric  Conditions  Determining  a  Line 

37.  The  Normal  Form 

38.  Reduction  of  a  Linear  Equation  to  the  Normal  Form 

39.  Distance  from  a  Line  to  a  Point        .... 

40.  Systems  of  Straight  Lines 

41.  Systems  of  Lines  through  the  Intersection  of  Two  Given 

Lines  ...... 

42.  Plotting  by  Factoring        . 


40 
40 
41 
41 
42 
44 
45 
47 
50 
50 
51 
54 
57 

59 
61 


CHAPTER   III  — THE   CIRCLE 

43.  The  Equation  of  the  Circle 

44.  The  General  Equation  of  the  Circle  .         .        .        . 

45.  Identification  of  the  Center  and  Radius   .        .        . 

46.  Special  Forms 

47.  The  Equation  of  the  Circle  Derived  from  Three  Con- 

ditions        ......... 

48.  Length  of  a  Tangent 

49.  Common  Chord  of  Two  Circles 

50.  Radical  Axis 

51.  Systems  of  Circles 


64 
66 
66 
67 

69 
70 

72 
73 
75 


CHAPTER   IV  — THE   PARABOLA 

52.  Conies         .... 

53.  The  Equation  of  the  Parabola 

54.  Discussion  of  the  Equation 

55.  Latus  Rectum    . 

56.  To  Draw  a  Parabola  . 

57.  Mechanical  Construction  of  the  Parabola 


77 
77 
78 
79 
79 
80 


CONTENTS 


Vll 


SECTION 

58.  The  Parabola  as  a  Conic  Section 

59.  The  Path  of  a  Projectile    . 


PAGE 

82 
83 


CHAPTER   V  — THE   ELLIPSE 

60.  Fundamental  Constants     . 

61.  The  Equation  of  the  Ellipse 

62.  Discussion  of  the  Equation 

63.  Second  Focus  and  Directrix 

64.  Focal  Radii 

65.  Latus  Rectum     . 

66.  Ellipse  with  Foci  on  the  ?/-axis 

67.  The  Circle  as  a  Limiting  Form 

68.  Construction  of  an  Ellipse  by  Points 

69.  Mechanical  Construction  of  the  Ellipse 

70.  The  Ellipse  as  a  Section  of  a  Cone   . 


85 
86 
86 
87 
89 
90 
90 
91 
91 
93 
94 


CHAPTER   VI  — THE   HYPERBOLA 

71.  Constants 96 

72.  The  Equation  of  the  Hyperbola 97 

73.  Discussion  of  the  Equation 98 

74.  Second  Focus  and  Directrix 98 

75.  Latus  Rectum 99 

76.  Hyperbola  with  Foci  on  the  y-axis 100 

77.  Focal  Radii 101 

78.  Asymptotes 101 

79.  Conjugate  Hyperbolas        .......  102 

80.  Equilateral  Hyperbolas 104 

81.  Construction  of  a  Hyperbola  by  Points     ....  104 

82.  Mechanical  Construction  of  the  Hyperbola       .         .         .  105 

83.  The  Hyperbola  as  a  Conic  Section 106 


CHAPTER  VII  —  TRANSFOR:\LiTION  OF 
COORDINATES  AND  SIMPLIFICATION 
OF   EQUATIONS 


84.  Change  of  Axes 

85.  Formulas  of  Translation 


107 
107 


Vlll 


CONTENTS 


SECTION  PAGE 

86.  Formulas  of  Rotation 108 

87.  Application  to  the  Conies 110 

88.  Test  for  Axes  of  Symmetry Ill 

89.  Simplification  of  Equations  by  Translation    .         .         .  112 

90.  Discussion  of  the  Equation  Ax"^  -\-  Cy"^  +  Dx  -\-  Ey  +  F  =  0  114 

91.  Limiting  Forms 116 

92.  General  Statement 116 

93.  Generalized  Standard  Equations  of  the  Conies       .         .  117 

94.  Simplification  of  Equations  by  Rotation         .        .         .119 

95.  The  General  Equation  of  the  Second  Degree  .         .121 

96.  The  Characteristic 121 

97.  Test  for  Distinguishing  the  Conies          ....  122 

98.  Suggestions  for  Simplifying  the  Equation  of  a  Conic     .  128 

99.  The  Conic  through  Five  Points 124 


CHAPTER   VIII  — POLAR   COORDINATES 


100.  Definition 

101.  Relations  between  Rectangular  and  Polar  Coordinates 

102.  Polar  Curves 

103.  The  Equation  of  the  Straight  Line 

104.  The  Equation  of  the  Circle 

105.  Discussion  of  Polar  Curves 

106.  Rotation  of  Axes 

107.  Symmetrical  Transformations 

108.  The  Equation  of  the  Conic 

109.  Equations  Derived  in  Polar  Coordinates 


127 
128 
129 
131 
131 
132 
136 
137 
138 
140 


CHAPTER   IX  — HIGHER   PLANE   CURVES 

110.  Algebraic  and  Transcendental  Equations 

111.  The  Exponential  Curve 

112.  The  Logarithmic  Curve 

113.  Applications     . 

114.  Periodic  Functions 

115.  The  Sine  Curve 

116.  Circular  Measure 

117.  The  Cosine  Curve 

118.  The  Tangent  Curve 


143 
143 
144 
145 
147 
147 
149 
149 
150 


CONTEXTS 


IX 


119.  Multiple  Angles       .... 

120.  Sum  of  Functions     .... 

121.  Inverse  Trigonometric  Functions    . 

122.  Parametric  Equations 

123.  Parametric  Equations  of  the  Circle 

124.  Parametric  Equations  of  the  Ellipse 

125.  The  Path  of  a  Projectile 

126.  The  Cycloid      .... 

127.  The  Epicycloid  and  Hypocycloid 

128.  The  Involute  of  the  Circle 

129.  Algebraic  Curves 

130.  The  Witch 

131.  The  Strophoid 

132.  The  Ovals  of  Cassini 


PAG» 

150 
151 
152 
153 
154 
154 
155 
157 
159 
161 
162 
162 
163 
164 


CHAPTER   X  — TAXGEXTS   AX^D   XORMALS 

133.  Definitions 

134.  Slope  of  Tangent  and  Xormal  to  the  Circle  at  a  Given 


Point 


135. 


Slope  of  Tangent  and  Xormal  to  the  Ellipse  at  a  Given 
Point 

136.  Slope  of  Tangent  and  Xormal  to  the  Parabola  and  Hy- 

perbola at  a  Given  Point  .         .         . 

137.  Equations  of  Tangent  and  Xormal  at  a  Given  Point      . 

138.  Tangent  and  Xormal.     Subtangent  and  Subnormal 

139.  Tangent  having  a  Given  Slope 

140.  Tangent  from  a  Given  External  Point    .... 

141.  The  Parabolic  Reflector 

142.  Angle  between  the  Focal  Radii  and  the  Tangent  to  an 

Ellipse  at  Any  Point 

143.  Diameters  of  a  Conic 

144.  Conjugate  Diameters  of  the  Ellipse         .... 

145.  Conjugate  Diameters  of  the  Hyperbola  .... 


166 

166 

167 

168 
169 
171 
171 
172 
174 

175 
176 
178 
179 


CHAPTER  XI  — SOLID  ANALYTIC  GEOMETRY 

146.  Introduction 181 

147.  Coordinates      ......•••     1^1 

148.  Radius  Vector  and  Direction  Cosines      ....     182 


CONTENTS 


150. 
151. 
152. 
153. 
154. 


SECTION 

149.   Distance  between  Two  Points 

Direction  of  a  Line  .... 

Angle  between  Two  Lines 

The  Locus  in  Solid  Geometry 

The  Normal  Equation  of  the  Plane 

Planes  Parallel  to  One  or  More  Coordinate  A 

155.  The  General  Equation  of  the  First  Degree 

156.  Angle  between  Two  Planes 

157.  The  Intercept  Equation  of  the  Plane 

158.  The  Equations  of  the  Line 

159.  The  Symmetrical  Equations    . 

160.  The  Two  Point  Equations 

161.  The  Projection  Forms 

162.  Reduction  of  the  Equations  of  a  Line  to  the 

Forms 

163.  Cylindrical  Surfaces 

164.  Surfaces  of  Revolution     . 

165.  Discussion  of  Surfaces 

166.  Quadric  Surfaces      .... 

167.  The  Ellipsoid 

168.  The  Hyperboloid  of  One  Sheet 

169.  The  Hyperboloid  of  Two  Sheets      . 

170.  The  Elliptic  Paraboloid  . 

171.  The  Hyperbolic  Paraboloid 
Formulas  and  Equations 


xes 


Standard 


PAGE 

183 
184 
185 
187 
187 
188 
189 
190 
190 
192 
192 
193 
193 

194 
196 
197 
199 
200 
200 
202 
203 
204 
204 
206 


ANALYTIC   GEOMETRY 


3     ^    \5       ' 


AI^ALYTIC   GEOMETRY 


INTRODUCTION 

DEFINITIONS    AND    FORMULAS    FROM    ALGEBRA    AND 

TRIGONOMETRY 

1.  The  Quadratic.  —  An  expression  reducible  to  the  form 

Ax"^  -{-  Bx+  C, 

where  A,  B,  and  C  are  any  constants,  is  called  a  quadratic 
in  X. 

2.  Solution  of  the  Quadratic.  —  The  roots  of  any  quadratic 
equation  may  be  found  by  completing  the  square,  or  by 
using  the  quadratic  formula. 

(a)  To  solve  by  completing  the  square,  transpose  the 
constant  term  to  the  right-hand  member  and  divide  both 
sides  of  the  equation  by  the  coefficient  of  x"^.  In  the  re- 
sulting equation  add  to  both  members  the  square  of  half 
the  coefficient  of  x  and  extract  the  square  root. 

(b)  If  the  above  rule  is  applied  to  the  general  quadratic 
Ax"^  +  Bx  -{-  C  =  0,  we  get 


^      -B±s/&-4.AC 
2A 

which  is  known  as  the  quadratic  formula.  To  solve  an 
equation  by  means  of  this  formula  simply  substitute  for  A, 
B,  and  C  in  the  formula  their  values  in  the  given  equation. 

3.   The  Discriminant.  —  The  expression  under  the  radical, 
B^  ^  4:  AC,  is  called  the  discriminant  oi  the  quadratic.     It 

1 


2  ANALYTIC   GEOMETRY 

is  denoted  by  the  G)eek  letter  A.     The  quadratic  formula 
shows  that  the  roots  of  a  quadratic  equation  are 

imaginary,  if  A  <  0 ; 

real  and  equal,       if  A  =  0 ; 
real  and  unequal,  if  A  >  0. 

4.  Logarithms.  —  The  logarithm  of  a  number  to  a  given 
base  is  that  exponent  by  which  the  base  must  be  affected 
in  order  that  the  result  may  be  equal  to  the  number.  Thus 
if  6^  =  N,  we  say  x  is  the  logarithm  of  N  to  the  base  b  and 
write  X  =  logj,  JSf. 

The  base  b  may  be  any  positive  number  different  from 
0  and  1.  Two  systems  of  logarithms  are  in  common  use: 
the  natural,  or  Naperian,  where  the  base  is  e  =  2.71828  .  .  .  ; 
and  the  common,  or  Briggs,  where  the  base  is  10. 

Important  properties  of  logarithms  to  any  base  are  the 
following : 

log,l  =  0;  log,6  =  l; 

M 

log,  MN=  log,  31  +  log,  N;       log6^=  ^^g,  M-  log,  AT; 

log,  N^=k  log,  N;  log,^i^=  ^  log,  N) 

log,A^  =  ^^g^=log,A^.log,6. 

5.  Angles.  —  In  trigonometry  an  angle  is  supposed  to  be 
generated  by  revolving  a  line,  called  the  generating  Ime, 
about  the  vertex  from  one  side  of  the  angle,  called  the 
initial  line,  to  the  other  side,  called  the  terminal  line.  If 
the  rotation  is  counter-clockwise,  the  angle  is  called  posi- 
tive ;  if  clockwise,  it  is  called  negative. 

Degree  Measure.  —  The  ordinary  unit  of  angle  measure- 
ment is  one  ninetieth  part  of  a  right  angle,  which  is  called 
a  degree.  One  sixtieth  part  of  a  degree  is  called  a  minute 
and  one  sixtieth  part  of  a  minute  is  called  a  second.     Thus 


INTRODUCTION  3 

60"  =  one  minute  ; 
60'  =  one  degree  ; 
360°  =  one  complete  revolution  =  four  right  angles. 
Circular  Measure.  —  The  unit  of  circular  measure  is  a 
central  angle  subtended  by  an  arc  equal  to  the  radius  of 
the  circle.     It  is  called  a  radian.     We  find  at  once  that 
IT  radians  =  180° ; 

1  radian  = : 


1°  =  -— -  radians. 
180 

From  this  definition  we  have  the  important  result  that  in 
any  circle  the  length  of  an  arc  equals  the  product  of  the 
measure  of  its  subtended  central  angle  in  radians  and  the 
length  of  the  radius. 

6.  Trigonometric  Functions.  —  From  any  point  P  of  the 
terminal  line  VP  of  the  angle  0  =  A  VP  drop  a  perpendic- 
ular on  the  initial  Ime  of  the  angle  (produced  if  necessary). 
The  ratios  of  the  sides  of  the  triangle  thus  formed  are 
called  the  trigonometric  ratios  or  functions  of  the  given 
angle. 


^/^ 

~v 

c 

A 

M 

k. 

^sT 

A 

Vv 

V^ 

i 

< 

ANALYTIC   GEOMETRY 


They  are 


.    .      MP  .      VM  .      ^     MP 

sm  6  = ;  cos  6  = ;  tan  6  = ; 

VP  VP  VM' 

G^(iO  = ;  sec^=-— -:  cot  »  =  -——. 

MP  VM'  MP 

In  these  ratios  the  side  of  the  triangle  opposite  the 
angle  is  considered  positive  when  it  extends  up  from  the 
initial  line,  negative  when  down;  the  adjacent  side  is  con- 
sidered positive  if  it  extends  to  the  right  of  the  vertex, 
negative  if  to  the  left ;  the  hypothenuse  is  always  positive. 


7.   Relations  between  the  Functions.  —  Eor  any  angle, 

/,        1  .  /J      cos  ^ 

sec  6  = ;  cot  ^  =  — 


cos  ^'  sin^' 

1 
sin  6 

1 

tan^ 

sin_^, 
cos  d' 


CSC  6  =  -^  ;  sin''  6  +  cos^  ^  =  1 ; 


cot  e  =  7-^ ;  1  +  tan2  $  =  sec^  6  •, 


tan^  =  ^^:  1  +  cot^  $  =  csc^ 0. 


8.   Reduction  of  Angles.  — 
sin  (— ^)  =  — sin.^;  sin  C|±^J=cos^; 

cos  (-  &)=      cos  e ;  cos  f'^±o\=T  sin  $ ; 

tan  (-^)  =  -tan^;  tan  f^±e\=Tcote; 

sin  (it  ±0)=T  sinO] 
cos  (tt  ±  ^)  =  —  cos  ^ ; 
tan  (tt  ±  ^)  =  ±  tan  6. 


INTRODUCTION 
9.   Special  Angles.  — 


Angle 

0 

TT 

TT 

TT 

T 

TT 

37r 

2ir 

6 

4 

3 

2 

2 

Sine 

0 

1 

V2 

V3 

1 

0 

—  1 

0 

2 

2 

2 

Cosine 

1 

V3 

2 

V2 
2 

1 
2 

0 

- 1 

0 

1 

Tangent 

0 

V3 

3 

1 

V3 

00 

0 

00 

0 

10.  Formulas  for  the  Sum  and  Difference  of  Two  Angles. 

sin  (6  ±  <f>)=  sill  0  cos  </>  ±  cos  ^  sin  <;^  ; 
cos  (^  ±  <^)  =  cos  0  cos  <^  qp  sin  ^  sin  <^ ; 

tan(^±<^)^  tan^±tanc^^ 
^     1 T  tan  ^  tan  <^ 

11.  Double  and  Half  Angle  Formulas.  — • 
sin  2  0  =  2  sin  6  cos  6', 

cos  2  ^  =  cos2  0  -  sin2  6*  =  1  -  2  sin^  0  =  2  cos^  ^  -  1  ?^ 
-  2  tan  ^ 


tan  2  ^  = 


sui 


0 


1  -  tan2  ^ ' 
—  cos  0 


tan  - 

2 


/I  -  cos  0  0        /I  -h  c 

=\/ ;  cos-=\/ — - — 

\        2  2^2 

-4 


cos  0 


—  cos  0     1  —  cos  0 


sin  ^ 


+  cos  0 


sin  0         1  +  cos  ^ 


12.   Triangle  Formulas.  —  In  an}^  triangle  ABC, 

— = —  = (Law  of  Sines)  ; 

sm  A      SYD.B     sm  C 

a^  =  b-  -\-  c"^  —  2  be  cos  A     (Law  of  Cosines). 


6 


ANALYTIC   GEOMETRY 


13.   The  Greek  Alphabet  — 

Letters 

Names 

Letters 

Names 

A,   a 

Alpha 

N,  V 

Nu 

B,  p 

Beta 

H,  1 

Xi 

r,  y 

Gamma 

O,    o 

Omicron 

A,    8 

Delta 

n,  TT 

Pi 

E,    c 

Epsilon 

P,    9 

Eho 

z,  C 

Zeta 

2,  ^ 

Sigma 

H,   rj 

Eta 

T,    r 

Tau 

@,  e 

•     Theta 

Y,  V 

Upsilon 

I,  t 

Iota 

$,  (^ 

Phi 

K,     K 

Kappa 

X,  X 

Chi 

A,  X 

Lambda 

^,  «A 

Psi 

M, /a 

Mu 

n,  w 

Omega 

CHAPTER   I 
COORDINATES  AND   EQUATIONS 

1.  Introduction.  —  The  chief  feature  of  Analytic  Geom- 
etry, which  distinguishes  it  from  the  geometry  which  the 
student  has  hitherto  studied,  is  its  extensive  use  of  algebraic 
methods  in  the  solution  of  geometric  problems.  Just  as  the 
use  of  symbols  in  algebra  makes  possible  the  ready  solution 
of  many  problems  which  would  be  difficult  if  not  impos- 
sible by  the  j^rocesses  of  arithmetic,  so  the  use  of  algebraic 
reasoning  simplifies  much  of  geometry  and  widens  its  scope. 
The  student  is  already  familiar  with  some  of  these  applica- 
tions ;  for  example,  the  theorems  stating  the  numerical 
properties  of  a  triangle  are  derived  in  whole  or  part  by 
algebraic  reasoning. 

In  trigonometry  also  the  greater  part  of  the  reasoning  is 
of  the  same  character,  and  further  simplification  is  gained 
by  the  introduction  of  negative  numbers.  For  example,  the 
law  of  cosines,  a^  =  6^  -^  c-  —  2  6c  cos  A,  states  in  compact 
language  the  contents  of  three  theorems  of  plane  geometry, 
since  cosine  A  is  positive  or  negative  according  as  angle  A 
is  acute  or  obtuse.  Other  examples  of  a  similar  character 
may  occur  to  the  student. 

2.  Position  ;  Cartesian  Coordinates.  —  The  most  important 
help  in  the  extension  of  geometric  analysis  has  been  found 
in  the  location  of  points  in  a  plane  or  in  space  by  Cartesian 
coordinates,  an  adaptation  to  mathematics  of  the  means  of 
expressing  position  used  in  ordinary  life. 

AYe  ordinarily  locate  a  point  by  stating  its  direction  and 
distance  from  a  fixed  reference  point,  as  —  Philadelphia  is 

7 


8 


ANALYTIC   GEOMETRY 


--5 


X' 


I     I     I     I     I 


-5 


A   r, 


X 


B 


P{4,-3) 


ninety  miles  southwest  of  Xew  York.  Less  commonly,  its 
position  is  fixed  by  its  distances  from  a  pair  of  fixed  per- 
pendicular lines.  An  illustration  of  this  is  found  in  the 
familiar  direction  given  in  a  city  to  go,  for  example,  eight 
blocks  east  and  three  north  to  reach  a  desired  destination. 
Both  of  these  methods  are  used  analytically  ;  the  second  is 
simpler  and  T^ill  be  discussed  first. 

In  the  figure  the  reference  lines  are  the  perpendiculars 
^'Xand  yy.     These  are  called  the  axes  of  coordinates  or 

coordinate  axes,  and  their 
intersection  0  is  called  the 
origin.  The  two  axes  di- 
vide the  plane  into  four 
quadrants  numljered  as  in 
trigonometry.  The  position 
of  a  point  P  is  determined 
by  measuring  its  distance 
from  Y'  Y  along  a  parallel 
to  X'X,  and  its  distance 
from  X'X  along  a  parallel 
to  Y'Y.  Distances  meas- 
ured to  the  right  of  I"'  l"or  up  from  X'X  are  called  positive^ 
those  to  the  left  or  down  negative.  The  measures  of  these 
distances,  with  the  proper  signs  prefixed,  are  called  the 
coordinates  of  the  point :  the  one  measured  along  a  line 
parallel  to  X'X  is  the  j:-co6rdinate  or  abscissa ;  the  one 
parallel  to  Y'Y  is  the  z/-coordinate  or  ordinate.  Thus  in 
the  figure  the  abscissa  of  P  is  BP=  -|- 4,  its  ordinate  is 
AP  =  —  3.  The  coordinate  axis  X'X  is  called  the  axis  of 
abscissas  or  A:-axis,  and  the  axis  Y'Y  the  axis  of  ordinate  s  or 
z/-axis. 

3.  Notation.  —  In  naming  a  point  by  its  coordinates, 
write  them  in  a  parenthesis,  putting  the  abscissa  first.  Thus 
in  the  figure  Pis  the  point  (4,  —  3).     In  case  the  coordinates 


-5 


COORDINATES  AND   EQUATIONS  9 

are  variable  or  unknown,  the  abscissa  is  denoted  by  the 
letter  x,  and  the  ordinate  by  y.  Fixed  points  of  which  the 
coordinates  are  not  known  or  are  arbitrary,  will  be  dis- 
tinguished by  means  of  subscripts,  being  lettered  Pi,  Pg,  etc., 
and  represented  by  the  coordinates  (xi,  y^,  {x2,  2/2),  etc. 

4.  Plotting  of  Points.  —  It  will  be  seen  that  any  point  in 
a  plane  is  fixed  by  means  of  its  coordinates  ;  for  the  abscissa 
locates  it  on  a  parallel  to  the  ^/-axis  and  the  ordinate  on  a 
parallel  to  the  a;-axis,  and  these  meet  in  one  point. 

To  locate  a  point  whose  coordinates  are  given  (or  as 
commonly  stated,  to  plot  the  point),  first  choose  a  convenient 
unit  of  measure,  then  measure  off  the  abscissa  on  the  a;-axis 


—(4  4) 

C-'?  '^^j- 

0 

- 

X31-2-)- 

A' 


Y' 


and  the  ordinate  from  the  end  of  the  abscissa.  Thus  to  plot 
(—3,  2)  count  3  unit's  to  the  left  on  the  a>axis  and  2  units 
up.  This  is  especially  convenient  when  using  coordinate 
paper,  i.e.  paper  ruled  with  sets  of  parallel  lines. 

The  system  of  coordinates  which  has  been  described  is 
the  rectangular  system  and  is  a  particular  case  of  Cartesian 
coordinates.  In  the  general  Cartesian  system  the  axes  are 
not  necessarily  perpendicular,  but  may  be  oblique.  All  of 
the  above  definitions  hold  for. the  oblique  system,  which  is 
illustrated  above. 


10  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Plot  accurately  the  points  : 

(a)   (2,5),(3,0),  (-V2,5)|^ 
(6)   (-3,0),(5,  -8),(1,V12); 

(c)  (2,  -6),  (-3,-3),  (0,  -Vt); 

(d)  (-1.2),  (-2, -3),  (-V3,0). 

2.  Draw  the  triangle  whose  vertices  are  as  follows  : 

{a)   (3,-1),  (-2,5),  (-8,  -4); 

(6)   (2,4),  (4,  V8)^(3,  11);  _ 

(c)   (5,  3),  (5  +  4V2,  3  +  4V2),  (5  +  8V2,  3  -  8  V2). 

3.  Draw  the  quadrilateral  whose  vertices  are  as  follows : 

(a)   (-2,0),  (2,4),  (6,0),  (2,  -4); 
(5)    (0,  -6),  (8,  2),  (0.  6),  (-8,  2). 

4.  Construct  with  compasses  on  coordinate  paper  V^.  V3,  Vo, 
and  estimate  the  value  of  each. 

Hint.  —  Each  radical  will  be  the  measure  of  the  hypothenuse  or  leg  of 
a  right  triangle. 

5.  Draw  a  circle  of  radius  10  with  its  center  at  the  origin.  Draw 
radii  at  20^  intervals  up  to  180^.  Estimate  from  the  figure  the  coordi- 
nates of  the  ends  of  these  radii. 

6.  From  the  figure  of  problem  5  make  a  table  of  sines  and  cosines 
of  the  angles  drawn. 

7.  What  is  the  abscissa  of  all  points  on  the  y-axis  ?  the  ordinate 
of  all  points  on  the  a;-axis  ?     What  are  the  coordinates  of  the  origin  ? 

8.  To  what  quadrants  Is  a  point  limited  if  its  abscissa  and  ordinate 
have  like  signs  ?  unlike  signs  ? 

9.  If  a  point  moves  on  a  parallel  to  the  x-axis,  which  of  its  coordi- 
nates remains  constant  ?  which  if  it  moves  on  a  perpendicular  to  the 
a:-axis  ? 

10.  (a)  What  is  the  locus  of  a  point  whose  abscissa  is  6  ?  whose 
ordinate  is  —  6  ? 

(b)  What  is  the  locus  of  all  points  having  the  same  abscissa? 
having  the  same  ordinate  ? 

11.  What  is  the  locus  of  points  whose  abscissas  and  ordinates  are 
(a)  equal  ?   (6)  numerically  equal,  but  of  unlike  sign  ?     Why  ? 

'•   12.    A  square  whose  side  is  2  a  has  its  center  at  the  origin  and  its 
sides  parallel  to  the  axes.     Eind  the  coordinates  of  its  vertices. 


COORDINATES  AND  EQUATIONS  11 

V  13.  An  equilateral  triangle  of  side  a  has  one  vertex  at  the  origin 
and  the  opposite  side  parallel  to  the  ?/-axis.     Find  the  coordinates  of 

the  other  vertices.  ^^^^     ^^  V3,  ±  ^) ,  or  (-  |  V3,  ±  f) ' 

-^  14.  A  rhombus  has  one  angle  of  60^  and  two  vertices  at  (0,  0)  and 
(a,  0).  Find  the  coordinates  of  the  other  vertices  if  (a)  both  are  in 
the  first  quadrant;  (&)  one  is  in  the  second  quadrant. 

.       ,  .     (a    aV3\    /3a   aV3\ 

•si  15.  A  regular  hexagon  of  side  a  is  placed  so  that  one  diagonal  lies 
along  the  x-axis  and  the  center  is  at  the  origin.  Find  the  coordinates 
of  the  vertices. 

5.  Directed  Lines.  —  We  have  referred  to  tlie  advantage 
of  using  positive  and  negative  lines  in  trigonometry  and 
have  defined  the  signs  of  coordinates.  In  analytic  geometry 
we  constantly  use  directed  lines,  that  is,  lines  whose  lengths 
are  reckoned  as  positive  or  negative  according  to  the  direc- 
tion in  which  they  are  read.  For  example,  if  the  positive 
direction  is  from  left  to  right,  and  AB  is  8  units  long,  then 
A B 

AB  =  8,  while  BA  =  —  8.  Thus,  changing  the  direction  of 
reading  a  line  changes  its  sign,  i.e.  BA  =  —  AB. 

In  adding  or  subtracting  line  segments  great  care  should 
be  taken  to  avoid  errors  of  sign.  It  is  advisable  for  begin- 
ners to  read  all  segments  in  one  direction  in  performing 
such  operations,  later  reversing  segments  as  required.  For 
example,  let  us  find  the  relation  between  AB,  BC,  and  AC 
in  the  figure. 

A C B 

We  have  AB=AC-{-CB  =  AC-  BC. 

In  the  Cartesian  coordinate  system  the  positive  direction 
on  all  lines  parallel  to  the  x-axis  is  to  the  right;  on  lines  not 
parallel  to  the  x-axis  the  positive  direction  is  upward.  All 
lines  should  be  read  m  the  positive  direction  unless  it  is 


12 


ANALYTIC   GEOMETRY 


intended  that  their  lengths  are  to  be  considered  negative, 
Since  coordinates  are  the  measures  of  directed  lines,  they 
must  be  read  up  or  to  the  right  if  positive,  and  doAvn  or  to 
the  left  if  negative. 

Exercise  1.  Prove  that  for  any  position  of  the  point  O  on  a 
directed  line  passing  through  the  points  A  and  B, 

AB  =  AC+  CB.  (Three  cases.) 

Exercise  2.  Prove  that  the  distance  between  any  two  points  hav- 
ing the  same  ordinate  is  the  difference  of  their  abscissas,  or 

d  =  xi  —  X2.  (Three  cases. ) 

M 

6.  Distance  Formula.  —  The  distance  between  two  points, 
Pi  (a^i,  y\)  <^f^nd  Pj  (^2j  yi),  li'liose  coordinates  are  known  is  given 
by  the  formula. 


For 
But 

and 


P^M=^  P^H^-  HM=  OK-  OB  =  x^  -  x^, 
MP,  =  MK^  KP,  =  KP,  -  BP,  =  y,-  y,. 


Substituting,  we  have  the  formula. 


Note.  — In  this  demonstration  and  those  which  follow,  the  fact  that 
the  formulas  are  true  for  all  positions  of  the  points  and  lines  involved  is 
of  fundamental  importance.  The  student  should  draw  the  figure  for 
several  possible  positions  and  satisfy  himself  that  the  demonstration  covers 
all  cases. 


COORDINATES  AND  EQUATIONS 


13 


7.  Point  of  Division  Formulas.  —  Tlie  coordinates  {xq,  2/0) 
of  a  poi7it  Pq  dividing  a  line  joining  Pi(x^,  y^)  and  Pgfe  ^2) 
in  a  given  ratio  r^ :  r^  are  given  by  the  formidas 


_  ^2X1  4-  r,x. 

Xq ^ ) 


ri4-r 


Hq ; 

''1+12 


(2) 


In  either  figure,  we  have,  by  similar  triangles, 

P,P,:P,P,  =  P,E:P,K 

But  PiPo :  AA  =  n:ro  by  hypothesis.     Also  P^E  =  OB 
-  0A  =  .To  -  x„  and  PoF=  OC-OB^x^-  Xq. 


Substituting, 
Solving  for  Xq, 


'  1  _  '<^0 


Xn  -^l 


JCf)   ~~~'    »//n 


^'1  +  ^2 


We  derive  the  formula  for  ?/o  from  the  triangles  PqFPj 
and  PiEPq  in  the  same  manner. 

When  Pq  is  the  mid-point  of  the  line,  r^  =  r^  and  these 
formulas  become : 


which  are  called  the  mid-point  formidas. 


(2  a) 


Exercise  3.     Derive  the  distance  formula  when  ,'    ^-  M^ 

}■  W(ci')  Pi  lies  in  the  fourth  quadrant  and  P2  in  the  second  ;  "^'.^ 


lb)  Pi  lies  in  the  second  quadrant  and  Po  in  the  third  ; 
(c)   Pi  lies  in  the  third  quadrant  and  P2  in  the  fourth. 


V^' 


14 


ANALYTIC   GEOMETRY 


Exercise  4.     State  the  rule  expressed  by  (a)  the  distance  formula, 
(6)  the  mid-point  formula. 

Exercise  5.     Derive  the  mid-point  formulas  directly  from  a  figure. 


^v         Exercise  6.     Derive  the  point  of  division  formulas  for  Pq  on  P1P2 
produced. 

Hint.  —  Note  the  sign  of  ri :  r2.  -;  ■'  ^         y\\  '    ■ 


8.  Geometric  Applications.  —  By  means  of  these  formulas 
and  others  which  will  follow,  many  of  the  theorems  of  plane 
geometry  can  he  proved  in  a  very  simple  manner.     '^     --■' 

Example.  —  Show  by  analytic 
means  that  the  mid-point  of  the 
hypothenuse  of  any  right  triangle  is 
equidistant  from  the  vertices. 

Solution.  —  Let  the  lengths  of  the 
legs  be  a  and  b.  Place  the  right  angle 
at  the  intersection  of  the  coordinate 
axes.  Then  the  vertices  are  (0,  0), 
(a,  0),  (0,  b).  Let  (7  be  the  mid-point 
of  the  hypothenuse.    By  formula  2  a 

its  coordinartes  are  ( -,-l.    The  dis- 


tance formula  gives 


O  '  '> 


The  same  result  is  obtained  for  CB  and  AC,  which  proves  the  theorem. 

In  this  problem  the  lengths  of  the  sides  were  expressed  by  letters 
so  as  to  insure  that  OB  A  should  represent  any  right  triangle.  The  justi- 
fication for  placing  the  right  angle  at  the  origin  lies  in  the  geometric 
axiom  that  figures  may  be  moved  about  in  space  without  altermg  their 
size  and  shape. 


PROBLEMS 
1.   Find  the  length  of  the  line  joining  (1,  3)  and  (-  2,  6). 

Solution.  — CM  (1,  3)  Pi  and  (-2,6)  Po.     Thenxi=l,  ?/i  =  3.  and  Xg  = 
—  2,  2/2  =  6.     Substituting  in  the  distance  formula, 

d  =  V(l  +  2)2  -t-  (3  -  6)2  =  V18  =  3v^. 


COORDINATES  AND   EQUATIONS  15 

2.   Find  the  lengths  of  the  sides  of  the  triangles  whose  vertices  are 

(a)   (4,3),  (2,  2),  (-3,  5); 
(6)    (-3,1),  (7,  -2),  (-6,5); 

(c)  (5,4),  (-3,2),  (-3,  -6); 

(d)  (4,5),  (2,-3),  (-6,  -3).  _       _       _ 

Ans.    (a)    V5,  V34,  V53. 

y'Z.    Show  that  the  following  are  the  vertices  of  isosceles  triangles : 

•wrJr-  («)  (2,4),  (5,1),  (6,5); 

'^  (&)  (2,6),  (6,2),  (-3,  -3). 

4.  Show  that  (3,  3),  (-3,  -3),  (3V3,  -3^3)  are  the  vertices 
of  an  equilateral  triangle,  and  find  the  length  of  its  medians. 

5.  Show  that  (3,  1),  (6,  5),  (  —  1,  4)  are  the  vertices  of  a  right  tri- 
angle. What  is  its  area  and  the  distance  between  the  mid-points  of 
its  legs  ? 

6.  Show  that  (15,  8),  (23,  -  7).  (8,  -  15),  (0,  0)  are  the  vertices 
of  a  square.     Find  the  lengths  of  its  diagonals. 

V  7.    Show  that  the  points  (3,  4),  (2,  1  -h  Vl2),  (4,  3)  lie  on  a  circle 
with  its  center  at  (1,  1). 

^^    >j   8.   Find  the  coordinates  of  the  point  dividing  in  the  ratio  3  :  5  the 
segment  whose  extremities  are  : 

(a)  (5,  6)  and  (13,  -10); 
(6)  (-  1,  5)  and  (3,  -9); 
(c)  (-6,  3)  and  (8,  -7).  .         Ans.    (a)   (8,0). 

9.  In  each  of  the  following  find  the  coordinates  of  the  point  three 
fourths  of  the  distance  from  the  first  to  the  second  point : 

(a)   (5,  6)  and  (13,  -  8) ; 
(6)   (-1,5)  and  (3,  -4); 

(c)   (-  6,  8)  and  (3,  -  2).    Ans.   (a) (11,  -  4.5). 
i|       Hint^^  Fi^7 determine  the  ratio  of  the  segments. 

/     ^  UO.   Find  the  point  of  intersection  of  the  medians  of  the  triangles 
^        whose  vertices  are  the  points  in  Problem  2.  Ans.    (a)   (1,  -V). 

ZZinf.  — We  know  from  plane  geometry  that  it  is  two  thirds  of  the  dis- 
tance from  a  vertex  to  the  middle  of  the  opposite  sidei 

11.  Find  a  point  10  units  distant  from  the  point  (—3,  6)  and  with 
the  abscissa  3. 

12.  Find  a  point  equidistant  from  the  points  (13,  8),  (6,  15)  and 
(-4,-9). 


/r^'- 


16  ANALYTIC   GEOMETRY 

18.  Two  of  the  vertices  of  an  equilateral  triangle  are  (1,  4)  and 
(3,  —  2).     Find  the  other  vertex. 

■■^  14.   Prove  analytically  that : 

(a)  the  diagonals  of  a  square  are  equal ;  ;^,.  >  "   -^"n^^iTi^ 

(b)  the  diagonals  of  a  rectangle  are  equal ;  '  ^  /" 
,  (c)  a  line  joining  the  middle  points  of  two  sides  of  a  triangle  is  one 

half  the  third  side  ; 

(d)  the  median  of  a  trapezoid  is  one  half  the  sum  of  the  bases. 

15.  Prove  analytically  that  in  any  triangle  : 

(a)  the  square  of  the  side  opposite  an  acute  angle  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides  decreased  by  twice  the 
product  of  one  of  those  sides  and  the  projection  of  the  other  upon  it ; 

(b)  the  sum  of  the  squares  of  two  sides  is  equal  to  twice  the  square 
of  one  half  the  third  side,  increased  by  twice  the  square  of  the  median 
on  that  side ; 

(c)  the  sum  of  the  squares  of  the  medians  is  equal  to  three  fourths 
the  sum  of  the  squares  of  the  three  sides. 

16.  Prove  analytically  that : 

(a)  the  sum  of  the  squares  of  the  four  sides  of  a  parallelogram  is 
equal  to  the  sum  of  the  squares  of  the  diagonals  ; 

(b)  the  sum  of  the  squares  of  the  four  sides  of  any  quadrilateral  is 
equal  to  the  sum  of  the  squares  of  the  diagonals  increased  by  four 
times  the  square  of  the  line  joining  the  mid-points  of  the  diagonals. 

9.  Angle  between  two  Lines.  —  Tlie  angle  of  intersection  of 
two  directed  lines  is  understood  to  he  the  angle  between  their 

positive  directions ;  i.e.  the 
angle  of  intersection  of  AB 
and  CD  is  BED.  There- 
fore in  Cartesian  geometry 
the  value  of  any  angle  lies 
between  0  and  tt.  For  ex- 
ample, suppose  the  line  CD 
to  be  rotated  around  E  so 
"  y^  ^^'^'  ^  that  the  angle  BED  in- 
creases from  0  until  ED  reaches  and  passes  the  position 
EA.  After  that  the  positive  direction  on  (7Z>  becomes  EC 
and  the  angle  between  the  lines  is  BEC. 


jj.-^' 


{C^' 


d^^. . 4--^  -"f*^ uot^Ti-.-^ 


COORDINATES  AND   EQUATIONS 


17 


10.  Inclination  and  Slope.  —  Tlie  inclination  of  a  line  is  its 
angle  of  intersection  with  the  x-axis.  If  it  is  parallel  to  the 
a>axis,  its  inclination  is  0. 

TJie  slope  of  a  line  is  the  tangent  of  its  inclination.  We 
denote  the  inclination  by  the  letter  a,  the  slope  by  m.  Thus 
?7i  =  tan  a.  When  m  is  positive,  a  is  acute ;  when  m  is 
negative,  a  is  obtuse,  and  conversely. 

The  slope  fixes  the  direction  of  a  line.  Thus  a  straight 
line  is  determined  when  its  slope  and  one  point  on  it  are 
known. 


^ 


5  -11.    The    Slope    Formula. 

—  The  slope  of  a  line  passing 
through  the  points  Pi(Xi,  y^) 
and  P2(^25  .^2)  *^  given  by 
the  formula 


9 


y 


m  — 


Xi  —  Xo 


(3) 


Take  the  case  where  Pi 
lies  in  the  first  quadrant 
and  Po  in  the  third. 


T 

y 

JMvPi) 

/ 

i' 

0 

K\        X 

PiiXo.lJn) 

H 

M 

Then 


KP,      MP, 
tan  «  =    .-f^  =  jTiTr- 
AR      P2M 


Now       MP,  =  MK  +  KP,  =  KPi  -  BP,  =  y,-  2/2, 
and  P23/=  P^H^  HM=  OK-  OB  =  x,  -  x^. 

Substituting,  we  have  the  formula. 

Exercise  7.     Derive  the  slope  formula  when 
(a)  Pi  is  in  the  fourth  quadrant  and  P2  in  the  second  ; 
(6)  Pi  is  in  the  second  quadrant  and  P2  in  the  third  ; 
(c)  Pi  is  in  the  third  quadrant  and  P2  in  the  fourth ;  . 
{d)  both  points   are  in  the   first  quadrant  and   the  inclination  is 
greater  than  90°. 

Exercise  8.     State  the  rule  expressed  by  the  slope  formula. 


18 


ANALYTIC   GEOMETRY 


12.  Parallelism  and  Perpendicularity.  —  If  tivo  lines  are 
parallel  their  slopes  are  equal,  and  conversely;  if  they  are 
perpendicular  their  slopes  are  negative  reciprocals,  and 
conversely. 


If  Zi  and   ^2   axe   parallel,  obviously  «,  =  a^,  whence  the 
slopes  are  equal. 

If  \  is  perpendicular  to  l^,  we  have  a-^  =  ac,-\-'^,  whence 


tan  «!  =  —  cot  ct2  =  — 


tan  tto 


This    gives    mi  = '-,    or 


??ln 


mi»?i2  =  —  1. 

Let  the  student  prove  the  converse  theorems. 
Summarizing,  we  have : 
Condition  for  parallelism,  jHi  =  ttZj  ; 

Condition  for  perpendicularity,     mjiti,  =  — 1. 


(4) 

(5) 


F 

yAtto 

\ 

\ 

^' 

0 
/ 

/ 

\^ 

13.   The    Angle    Formula. — 

Tlie  angle  between  two  lines  is 
given  by  the  formula 


tanp 


m^  —  m^ 
1  -{-m^m^ 


(6) 


X    m^    denoting   the    slope    of  the 
line  of  greater  inclination. 


COORDINATES  AND   EQUATIONS  19 

Obviously        a^  =  05  +  ^  and  fi  =a^—  a^, 

wheuce  tanfl=  tan  «,  -  tan  «, 

1  +  tan  «!  tan  Og 

_   TTli  —  7712 


1  +  mi??i2 


Exercise  9.     Derive  the  angle  formula  for  the  case  when  the  point 
of  intersection  is  below  the  a;-axis. 


PROBLEMS 

1.    Find  the  slope  of  the  line  joining 

(a)  (2,  3),  (3,  5)  ; 

(6)  (3,8),  (-6,  -Q)-, 

(c)  (V2,  V3),  (_  V3,  V2); 

(d)  (a  +  &,  c),  (a,  b-\-  c). 

^'  2.  What  is  the  inclination  of  the  line  joining 
(a)  (3,  3)  and  (-  3,  -  3)  ; 
(6)  (-3,  2)  and  (-4,  3); 

(c)  (5,  0)  and  (6,  \/3)  ;  _ 

(d)  (-3,0)  and  (_2,V3);. 

(e)  (0,  -4)  and  (-3,  1); 
(/)   (0,  0)  and  (-  V3,  -  5)  ? 


Ans.  (a)  45°. 


/jA-o 


3.  Prove  by  means  of  slopes  that  (2,  3),  (6,  —  3)  and  (—  2,  9)  are 
on  the  same  straight  line. 

4.  Prove  that  (4,  7)  is  on  the  line  joining  the  points  (6,  6)  and 
(2,  8),  and  is  equidistant  from  them. 

V^.  Prove  that  the  line  joining  (6,  -  3)  and  (2,  8)  is  perpendicular 
!b  the  one  joining  (0,  0)  and  (11,  4). 

6.    What  is  the  inclination  of  a  line  parallel  to  Y'Y?  perpendicu- 
lar to  Y'  Y?    What  are  the  slopes  of  these  lines  ? 

|/  7.   Find  the  angle  between 

y^  (i)  lines  (a)  and  (d)  in  Problem  2 ;     ' 
*''(ii)  lines  (&)  and  (e)  in  Problem  2  ; 
(iii)  lines  (c)  and  (/)  in  Problem  2. 

Ans,  (i)  15°. 


>i 


20  ANALYTIC    GEOMETRY 

JL-^  J^^^'   Find  the  angle  between  the  line  from  (4,  —  2)  to  (—  3,  6)  and 
e  line  bisecting  the  first  quadrant. 

9.    Prove  by  means  of  slopes  that  the  following  points  are  the 
vertices  of  a  right  triangle  and  find  its  angles  : 

(a)   (5,  0),  (8,  4),  (1,  3)  ; 
(6)   (5,  5),  (2,  8),  (10,  10). 

10.  Prove  by  means  of  slopes  that  the  following  points  are  the 
vertices  of  a  parallelogram  and  find  its  angles  : 

(a)    (-1,  -2),  (0,1),  (3,4),  (2,1); 
{h)  (3,  -3),  (4,0),  (7,3),  (6,0). 

11.  Prove  that  (0,  -  2),  (4,  2),  (0,  6),  and  (~  4,  2)  are  the  ver- 
tices of  a  square. 

J.  12.   Find  the  slopes  of  the  sides  and  the  angles  of  the  triangle 
whose  vertices  are  : 

(a)  (-  2,  2),  (4,  2),  (1,  5)  ;   :  (&)  (1,  1),  (-  1,  -  1),  (5,  -  5). 
j^  13.    Prove  analytically  that 

(a)  the  diagonals  of  any  square  are  perpendicular; 

(6)  the  median  of  a  trapezoid  is  parallel  to  the  bases  ; 

(c)  the  lines  joining  the  mid-points  of  the  sides  of  any  quadrilateral 
form  a  parallelogram ; 

{d)  the  lines  joining  the  mid-points  of  the  sides  of  a  rectangle  form 
a  rhombus. 

14.  What  is  the  relation  between  two  lines  for  w'hich  wii  =  —  m-i  ? 

15.  Three  vertices  of  a  parallelogram  are  (8,  6),  (2,  5),  and  (4,  —2). 
Find  the  fourth  vertex. 

16.  Find  a  point  on  the  line  joining  (2,  3)  and  (—4,  6)  and  2 
units  distant  from  the  latter  point. 

17.  Derive  formula  (5)  from  formula  (6). 

14.  Equations  and  Graphs. — The  coordinates  x,  y,  if  no 
restriction  is  placed  on  their  values,  represent  any  point  in 
the  plane.  If,  however,  the  values  of  x  and  y  are  subject 
to  certain  conditions,  points  having  these  coordinates  will 
lie  upon  certain  lines  or  curves.  For  examj)le,  if  y  is  unre- 
stricted, but  X  always  equals  —  6,  all  such  points  will  lie 
upon  a  line  parallel  to  the  ?/-axis  and  6  units  to  the  left. 
Again,  the  locus  of  all  points  whose  abscissas  and  ordinates 


COORDINATES  AND   EQUATIONS  21 

are  equal  is  the  bisector  of  the  first  and  third  quadrants. 
Such  restrictions  upon  coordinates  are  expressed  by  means 
of  equations.  Thus,  the  equation  of  the  first  locus  is 
x  =  —  G,  of  the  second  x  =  y.  As  a  general  definition,  we 
have: 

The  equation  of  a  locus  is.  an  equation  satisfied  by  the  coor- 
dinates of  all  points  lying  on  the  given  locus,  and  conversely. 
The  curve  which  contains  all  points  whose  coordinates  sat- 
isfy the  given  equation  and  no  other  points  is  called  the 
locus  or  graph  of  the  equation. 

The  derivation  of  equations  of  loci  and  the  study  of 
graphs  by  means  of  their  equations  form  the  chief  part  of 
elementary  Analytic  Geometry. 

15.  Plotting  of  Graphs.  — When  a  locus  is  given  by  its 
equation,  the  shape  of  the  curve  is  sometimes  evident  from 
the  form  of  the  equation,  as  in  examples  mentioned  above, 
and  as  we  shall  see  in  more  complicated  problems  later. 
Usually  the  graph  is  constructed  by  a  process  called  plot- 
tnig  the  graph  of  the  equation,  in  which  we  proceed  as 
follows  : 

Solve  the  equation  for  y  in  terms  of  x. 

Set  X  equal  to  convenient  positive  and  negative  values  (gen- 
erally integral)  and  compute  the  corresponding  values  of  y. 
Each  pair  of  values  of  x  and  ?/  is  a  solution  of  the  equation, 
and  hence  the  point  of  which  these  are  the  coordinates  lies 
on  the  curve  by  the  definition  of  the  locus. 

Make  a  table  of  values  by  arranging  these  pairs  in  order 
according  to  the  magnitudes  of  the  values  of  x. 

Plot  the  points  thus  tabidated  and  join  them  by  a  smooth 
curve  in  the  order  of  the  table.  This  gives  an  approximation 
to  the  true  curve  which  becomes  more  exact  when  a  larger 
number  of  points  is  plotted. 

Remarks.  — It  is  sometimes  more  convenient  to  solve  for  x  in  terms  of 
y.     In  this  case  the  above  is  applicable  on  interchanging  x  and  y. 


22 


ANALYTIC   GEOMETRY 


If  the  solution  for  y  in  terms  of  x  involves  a  square  root,  the  double 
sign  must  be  used  with  the  radical  and  in  general  two  points  will  be  found 
for  each  value  of  x.  A  similar  remark  applies  to  a  solution  for  x  in  terms 
oft/. 

Sometimes  the  values  of  a;  or  y  will  be  so  large  that  it  is  difl&cult  to 
draw  the  curve  on  the  plotting  paper.  In  this  ease  use  one  scale  for  x  and 
another  for  y. 

"When  the  points  plotted  do  not  show  the  form  of  the  curve,  look  for 
an  error  in  the  table  of  values.  If  there  is  no  error,  assume  intermediate 
fractional  values  of  x  or  y  and  plot  the  corresponding  points. 

The  table  of  values  should  be  extensive  enough  to  show  the  form  of  the 
curve  completely.  If  it  extends  to  infinity,  the  plot  should  contain  all 
parts  with  considerable  curvature  and  should  extend  far  enough  to  show 
the  direction  of  the  curve  beyond  the  limits  of  the  paper. 

^  Example  1.  —  Plot  the  graph  of  3  jC  -{-?/  —  4  =  0. 


Solving  for  ?/, 

Substituting    a:  =  —  2, 

—  1,  0,  1,  2,  3,    we   get 

2/ =  10,  7,4,  1,-2,   -5 

respectively.  Thus  points 

etc.  oil    the    line    are    (—  2, 

10),  (-1,7),  (0,  4),  etc. 

The  table  of  values  is  at  the  left  and 

the  plot  at  the  right. 


X 

y 

-2 

10 

-1 

7 

0 

4 

1 

1 

2 

-  2 

3 

—  5 

Example  2.  — Plot  the  graph  of  y-  —  2  y  =  :r  —  3. 

Here  it  is  easier  to  solve     Y  ' 
for  X  in  terms  of  y  : 

x  =  y'--2y  +  S. 

In  the  plot  each  unit  on 

the  ?/-axis   has   been   taken 

equal  to  two   units  on   the 

X-axis.     This  is  because  the 

values  of    x    are    so    much 

larger  than  those  of  y  that 

if  the  same  scale  were  used 
on  both  axes,  the  curve  would  be  hard 
to  draw.     In  such  cases  the  scale  used  should  be  indicated  on  the 
graph. 


y 

X 

-3 

18 

-2 

11 

-1 

6 

0 

3 

1 

2 

2 

3 

3 

6 

4 

11 

5 

18 

^- 


COORDINATES  AND   EQUATIONS 


23 


Q  L  'Example  3.  —Plot  the  graph  of  4  x^  +  y"- =  20. 


X 

y 

-3 

imag. 

-2 

±2 

-1 

±4 

0 

±  2V5 

=  ±4.5 

1 

±4 

2 

±2 

3 

imag. 

SoMn.o: 
of  X, 


for  ?/  in  terms 


y  =±  V20  —  4 x^ 
_,=  ±  2V5-r/:^. 

Here  no  part  of  the 
curve  can  lie  outside  of 
X  =  +  3  and  x  =  —  3, 
"Whereas  there  are  two 
points  corresponding  to 
X  =  ±  2.  To  find  out  where  the  curve 
crosses  the  x-axis,  we  set  ?/  =  0  in  the  equa- 
tion and  solve,  getting  x  =  ±  \/5=  ±  2.2+. 
Adding  the  points  (2.2+,  0)  and  (-  2.2+, 
0)  to  the  plot,  we  are  able  to  draw  the 
entire  curve. 


Example  4.  —  Plot  the  graph  of  xy  =  —  4,  or  y  — 


)..^ 


X 

y 

1 

-4 

2 

-2 

3 

-f 

4 

-  1 

6 

-1 

8 

_  1 
1. 

X 

y 

-  1 

4 

-2 

2 

-3 
-4 

f 

1 

-6 
-8 

1 

Since  x  cannot  equal  0*.  we 
must  take  fractional  values  of  x 
between  ±  1  in  order  to  find  the 
shape  of  the  curve.  These  give 
additional  points  : 


X 

y   ! 

X 

y 

i 

-  8    1 

-i 

8 

1 

-  16 

-i 

16 

Adding  these  to  the  plot,  we  see  that  the  curve  goes  to  infinity 
along  the  x-  and  2/-axes  in  the  second  and  fourth  quadrants. 


*  This  is  due  to  the  fact  that  the  substitution  of  x  =  0  in  the  equation 
involves  division  by  0,  an  impossible  operation.  Since  y  becomes  indefi- 
nitely great  as  x  approaches  0,  it  is  sometimes  stated  that  for  x  —  0,y  =  », 
but  this  abbreviation  should  not  be  allowed  to  obscure  the  fact  that  for 
a;  =  0  there  is  no  point  on  the  curve. 


24  ANALYTIC    GEOMETRY 


PROBLEMS 

Plot  the  graphs  of  the  following  equations  : 

1. 

X  =  6. 

J  11. 

9  r;2  +  2/2  =  36. 

2. 

y  =0. 

18. 

x2  +  2  2/2  =  2. 

3. 

y  =  -2. 

19. 

25  x2  +  16  ?/2  =  400. 

4. 

Sy  =  x. 

20. 

x2  -  ?/2  =  9. 

5. 

y=-2x. 

21. 

^2  =  25  +  x2. 

6. 

3x  +  4:y  =0. 

:  22. 

x2  +  4  X  +  4  =  ?/2  -  3  2/. 

4  X  -  3  ?/  =  6. 

23. 

x2  -  9i/2  =  36. 

8. 

x  +  y  =  1. 

24. 

9x2-2/2  +  9  =  0. 

9. 

?/2  =  X^. 

6  25. 

X2/  =  12. 

^10. 

y  =  x^  —  ox  -{-  4. 

26. 

2xy  =-  15. 

11. 

x^-Sx+2y  =  6. 

27. 

?/  =  X'^  —  9  x2. 

12. 

X  =  y^  —  6  y  ■}-  ^. 

28. 

2/^  =  x2. 

13. 

2/2  -  2/  =  X  +  2. 

29. 

xy  =  y  +  2. 

14. 

x^  +  y^=  16. 

30. 

xi/  =  3x  —  1. 

15. 

j,2  +  2/2  =  25. 

31. 

xy  -y^-  =  10.  (Solve  for  x.) 

16. 

x2  +  4  2/2  =  4. 

32. 

X  +  2/  —  2  x?/  =  0. 

33. 

Find  two  of  the  above 

curves  whicl 

1  pass  through 

(a)    (0,0);         (6)   (-1,  -^  1)  ;         (c)    (1,1). 

16.  Derivation  of  Equations.  —  The  derivation  of  equations 
of  loci  or  curves  is  a  process  depending  largely  upon  the 
ingenuity  of  the  solver  of  the  problem.  However,  the 
following  general  suggestions  will  be  helpful. 

1.  Take  the  origin  and  axes  in  a  convenient  position.  The 
origin  will  usually  be  a  fixed  point  mentioned  in  the  prob- 
lem, or  the  point  midway  between  two  such  points,  with  one 
of  the  axes  passing  through  them.  This  does  not  involve 
any  loss  of  generality  as  far  as  the  locus  is  concerned,  since 
the  locus  is  independent  of  its  position  in  the  plane ;  while  a 
correct  choice  of  axes  will  greatly  simplify  the  form  of  the 
equation  and  the  work  of  deriving  it. 

2.  Mark  P  (x,  y)  as  any  point  satisfying  the  given  condi- 
tions and  therefore  on  the  curve.  Draw  its  coordinates  if 
necessary. 

3.  Draic  any  line  suggested  by  the  data  of  the  problem. 


COORDINATES  AND   EQUATIONS 


25 


It  should  contain  all  the  given 


4.  Express  the  conditions  of  the  problem  in  an  equation  con- 
taining X,  y,  and  the  given  constants.  For  this  purpose  it 
will  be  necessary  to  lind  some  formula  or  principle  of 
geometry  relating  to  the  coordinates  and  given  constants. 

5.  Simplify  the  equation. 
numerical    or    arbitrary    con- 
stants, but  no  variables  other 
than  X  and  y. 

Example  1 .  —  Find  and  plot  the 
locus  of  points  distant  6  units  from 
the  point  (7,  —  4). 

SoIutio7i.  —  Here  the  coordinate 
axes  are  fixed  by  the  statement  of 
the  problem.  Take  any  point  P{x,  y) 
as  one  satisfying  the  conditions  of 
the  problem  and  draw  CP.  Since 
CP  has  the  constant  length  G,  the 
distance  formula  is  suggested.  This 
gives  at  once 

r)=V(x-7)2  4-(?/+4)2,  or  (x-7)2+(j/+4)2=36. 

We  can  either  plot  the  curve  by  points,  or,  since  we  know  that  it  is  a 
circle,  draw  it  with  compasses. 

^  Example  2. — A  rock  in  the  ocean  lies  6  miles  off  a  stretch  of 
straight  coast  and  a  ship  moves  so  as  to  be  always  equidistant  from 
the  rock  and  the  coast.     Find  the  locus  of  the  ship. 

Solution.  —  Take  the  coast  as  the  .r-axis  and  let  the  ?/-axis  pass  through 
the  rock,  which  will  have  the  coordinates  (0,  G).  Take  a  point  P(x,  y) 
and  draw  its  ordinate  ^IP.    Then  OA  =  x,  AP  =  y.    P  is  to  be  equidistant 


Y 

^'' 

P{x,y) 

R 

-{pfi) 

0 

i?^ 

i     X 

^'■/^-f^  ; 


26  ANALYTIC   GEOMETRY 

from  R  and  OA.  This  suggests  drawing  RP,  AP  being  already  drawn. 
By  the  conditions  of  the  problem  RP  =  AP.  To  get  the  length  of  RP  use 
the  distance  formula,  which  gives 

RP  =  Vx=^  +  (2/  — t))'^-    But  its  equal  AP  =  y 

.-.  Vx2  +  (?/  — 6)2=2/,  whence  a;2_  12  y +  36  =  0. 

The  derivation  of  the  above  equations  shows  that  the 
coordinates  of  all  points  which  lie  on  the  locus  satisfy 
the  equation  obtained.  To  show  that  the  equation  fulfills 
the  second  requirement  of  the  definition  of  the  locus, 
namely,  that  any  point  the  coordinates  of  which  satisfy  the 
equation,  lies  on  the  locus,  it  is  merely  necessary  to  see  if 
the  various  steps  taken  in  the  derivation  of  the  equation  can 
be  retraced.  To  illustrate,  in  Example  2  any  point  P{x,  y)i 
whose  coordinates  satisfy  the  equation  x^  —  12  2/  +  36  =  0, 
must  also  satisfy  the  previous  equation  since  it  can  be  re- 
duced to  that  form  by  reversmg  the  steps  by  which  the 
latter  was  obtained.  But  the  relation  •\/x^  +  (?/  —  6)2  =  ?/ 
simply  says  that  the  point  P  is  equidistant  from  the  a;-axis 
and  R,  which  was  the  condition  stated  in  the  problem. 

PROBLEMS 

1.    What  is  the  equation  of  a  straight  line 

(a)  parallel  to  the  cc-axis  and  6  units  above  it ; 
.  .     f\  (h)  parallel  to  the  ?/-axis  and  6  units  at  the  left  ? 

^  2.   Find  and  plot  the  equation  of  the  locus  of  points 

(a)  three  times,  (6)  m  times 

as  far  from  the  x-axis  as  from  the  y-axis. 

<  3.    Solve  Example  2,  page  25,  when 
.(a)  the  origin  is  at  J?  ; 
^  (6)  the  origin  is  midway  between  0  and  B. 

/,4.    Find  and  plot  the  equation  of  ±he  locus  of  points  equidistant 

f^«°^  K(«)   (3.^  ^^d  (6,  i  ^  ;  I7  ^ 

(6)    (-4,  3)  and  (6,  0); 


(c)   (-5,0)  and  (0,4); 
-^  ^       [  (.d)   (0,  0)  and  (a,  0). 


lO 


I 


II 


COORDINATES  AND   EQUATIONS       '  27 


/ 


lb.   Find  and  plot  the  equation  of  the  circle  having 
V  (a)  radius  3  and  center  (—  5,  6)  ; 
(&)  radius  6  and  center  (1,  —  3) ; 
(c)  radius  2  and  center  (0,  0)  ; 
(cZ)  radius  r  and  center  (^,  k).  '         ' 

6.    Find  and  plot  the  equation  of  the  locus  of  points  equidistant 
ftom  (^a)  the  line  y  -Q  and  the  point  (3,  0)  ; 

(6)  the  line  u;  =  4  and  the  point  ( -  2,  0)  ; 

(c)  the  line  x  =  —  8  and  the  point  (0,  —  5)  ; 

{d)  the  line  x  =  0  and  the  point  (p,  0).  ^ 

^/  7.    Find  the  locus  in  Problem  6  when  the  distance  from  the  point'  f 

is  twice  that  from  the  line. 

8.  Find  the  locus  in  Problem  6  when  the  distance  from  the  line  is 
twice  that  from  the  point. 

9.  Find  the  equation  of  the  straight  line  passing  through 
^'^'              {a)   (3,  4)  and  (5,  -  2)  ;  (c)   (1,  1)  and  ( -  2,  -  2)  ; 

(6)  (4,  3)  and  (-  1,  6)  ;  (d)   (a,  0)  and  (0,  h). 

\^  10.    Find  the  equation  of  the  straight  line  passing  through  (—  2,  3) 
and  of  inclination 

(«)   120^   (6)  ^;     (c)  150°;     (d)  ^. 

11.  The  distance  between  two  fixed  points  is  2  c.  Find  the  locus 
of  a  point  moving  so  that  the  sum  of  the  squares  of  its  distances  from 
the  points  is  4  c^. 

12.  The  ends  of  a  straight  line  of  variable  length  rest  on  two  per- 
pendicular lines.  Find  the  locus  of  the  middle  point  if  the  area  of  the 
triangle  formed  is  constant. 

^*"    13.    The  base  of  a  triangle  is  of  length  2  a.     Find  the  locus  of  the 
^j^vertex  if  the  vertical  angle  is  90°. 

"N  17.  Functional  Variables.  —  The  symbols  used  in  mathe- 
/  ^  matics  represent  two  kinds  of  quantities,  variables  and 
^  constants. 

A  variable  is  a  quantity  which  takes  on  an  unlimited  num- 
ber of  values  ;  e.g.  the  velocity  of  a  falling  body,  the  abscissa 
of  a  point  moving  along  a  curve,  etc.,  are  variable  quantities. 
A  constant  is  a  quantity  having  a  fixed  value.     There  are 
two  kinds  of  constants  :  absolute  constants,  which  have  the 


28  ANALYTIC    GEOMETRY 

same  value  in  all  problems,  as  2,  —  6,  tt  ;  and  arbitrary 
constants,  which  may  have  any  value  assigned,  but  keep 
the  same  value  in  a  given  discussion,  as  the  quantities  m^ 
and  77I2  in  the  discussion  of  parallel  and  perpendicular  lines 
in  §  12. 

When  two  variables  are  connected  by  some  law  such  that  the 
value  of  one  depends  upon  that  of  the  other,  the  first  variable 
is  said  to  be  a  function  of  the  other.  The  first  variable  is 
called  the  dependent  and  the  second  the  independent  variable. 
Thus  the  area  of  a  circle  is  a  function  of  the  radius,  the 
pressure  of  steam  is  a  function  of  the  temperature,  etc. 
Such  relations  are  commonly  expressed  by  means  of  equa- 
tions, as,  in  the  case  of  the  formula  expressing  the  area  of 
a  circle,  A  =  irr'^. 

18.  Functional  Notation.  —  The  expression  f{x)  is  used  to 
denote  any  function  of  x\  it  is  read  "function  of  a;,"  or 
"/  of  «."  Similarly  f(x,  y)  stands  for  a  function  of  both 
X  and  y.  To  denote  a  different  function  of  x,  some  other 
letter  is  used,  as  F(x)  or  g(x).  Since  functional  relations 
are  usually  expressed  by  means  of  equations,  the  expression 
f(x)  is  ordinarily  an  abbreviation  for  some  combination  of 
terms  containing  x  and  no  other  variable. 

If  an  equation  in  x  and  y  is  solved  for  y,  we  regard  x  as 
the  independent  variable  and  2/  as  a  function  of  x.  Thus, 
if  x  and  y  are  connected  by  the  relation  ic^ -f- 4  ?/2  =  4, 
y=±^ V4—  x~  may  be  denoted  by  the  abbreviation 
y=f(x),  where  f(x)  stands  for  the  quantity  ±-i-V4  — .'c^. 
The  symbols  /(I),  /(—  2)  represent  the  values  of  the  func- 
tion when  1  and  —2  respectively  are  substituted  for  x. 
For  this  function 

/(l)=:±iV4^^  =  ±iV3,/(-2)=±iV4^^  =  0. 

If  it  is  not  desired  to  solve  the  equation  for  either  variable, 
we  may  transpose  all  terms  to  the  first  member  and  denote 
this  form  of  the  equation  by  the  expression /(a;,  y)  =  0. 


COORDIXATES  AND   EQUATIONS 


29 


19.  Graphs  of  Functions.  —  If  we  plot  the  graph  of  an 
equation  expressing  a  functional  relation,  the  graph  may 
be  regarded  as  a  pictorial 
representation  of  the  relation 
between  the  variables.  Con- 
sider, for  example,  the  rela- 
tion between  the  velocity  of 
a  ball  thrown  vertically  up 
with  an  initial  velocity  of  64 
feet  per  second  and  the  height 
which  it  has  attained  at  any 
time.  By  a  physical  formula 
^1  =  642  _  64  h. 
Let  each  unit  on  the  .r-axis  represent  1  foot  and 
each  unit  on  the  ?/-axis  represent  a  velocity  of  1 
ft.  per  second.  For  each  value  of  h  the  correspond- 
ing ordinate  represents  the  velocity  of  the  ball  at 
that  height. 

In  presenting  statistical  information  the  graph  is  con- 
stantly used.  Here,  as  a  general  thing,  the  functional  rela- 
tions cannot  be  expressed  by  means  of  equations,  and  the 
graphs  are  plotted  from  observed  data.  The  following 
table  gives  the  debt  of  the  United  States  for  the  years  1875 
to  1910,  the  unit  being  $100,000,000,  and  the  adjoining 
graph  presents  the  same  mformation. 


"o'|64 
16  55 
32 '45 

48  32 
64    0 


Debt  i>* 

Year 

Hundreds  of 

MiLLTOXS 

1875 

22 

1880 

21 

1885 

19 

1890 

15 

1895 

17 

1900 

21 

1905 

23 

1910 

27 

1875     1885     1895     1905 


30  ANALYTIC   GEOMETRY 

PROBLEMS 

1.    Write  the  equation  2x^  +  y^  =  4:xy  in  the  three  forms  y  =f{x)^ 
a^=/(2/),  and/(x,  y)  =0. 

Solution.  —  To  express  y  in  terms  of  x,  solve  the  equation  for  y, 

■        2 
Hence  y  =f{x)  =  2  a:  ±  a;  V2. 

lu  like  manner      ^  =/(?/)  =  2/ db  i2/V2, 
and  f{x,y)=2x-^-^zy  +  y^  =  0. 

^^z.  "Write  each  of  the  following  equations  in  the  three  forms 
y  =f{x),  x=f{y),  and/(.r,  y)=0: 

(a)  xy  +  y  =  6;  ^c)  x^ -{- 2x  =  y'^  -  6y ; 

Xb)  oi'  +  9^2  ^  9;  {d)  y={x-\-  If. 

*y^.   If  f  (rr)  =  a;3  _  2  :e2  +  Sx,  find  the  value  of  /(I).  /(O),  /(-  3), 

4.  If /(x)  =  6x*  -  os;2  +  3,  show  that/(-  a:)  =/(x)  identically. 

5.  If /(a:,  2/)  =  8x2  +  3?/-  ?/2,  show  that /(-  x,  2/)  =/(x,  y)  iden- 
tically.    Does/(x,  -  ?/)  =/(:'•,  y)  ? 

6.  Write  a  function  of  x  and  2/  such  that 

(a)  f(x,  -  y)  =f(x,  y)  identically; 
(&)  /(-  X,  -  y)  =f(x,  y)  identically  ; 
{c)  f{-x,y)  =/(x,  y)  identically. 

7.  Express  the  radius  of  a  sphere  as  a  function  of  the  surface  and 
plot  the  graph  of  the  function. 

8.  The  area  of  a  triangle  of  variable  base  and  altitude  is  always 
16  square  feet.  Express  the  base  as  a  function  of  the  altitude  and 
plot  the  graph  of  this  function. 

9.  A  point  moves  around  the  circumference  of  a  circle  of  radius  5. 
Express  its  distance  from  one  end  of  a  fixed  diameter  as  a  function  of 
its  distance  from  the  other  end  and  plot  the  graph  of  this  function. 

:  10.  A  rectangle  of  varying  size  has  one  side  3  inches  longer  than 
the  other.  Express  the  area  in  terms  of  the  shorter  side  and  plot  the 
graph  of  this  function. 

11.    From  an  almanac  find  the  length  of  daylight  on  the  first  of  each 
month  during  the  year  and  draw  a  graph  illustrating  this  data. 

20.   Discussion  of  Equations.  —  Since  it  is  jDossible  to  plot 
but  a  few  points  on  a  curve,  the  graph  is  always  more  or 


COORDINATES  AND  EQUATIONS 


31 


less  inaccurate.  To  discover  the  properties  of  a  curve  it  is 
necessary  to  study,  or  discuss,  its  equation.  The  purpose 
of  this  discussion  is  threefold :  it  gives  exact  information  re- 
garding the  curve,  it  furnishes  a  check  upon  the  accuracy 
of  the  plot,  and  it  usually  facilitates  the  labor  of  plotting. 
The  properties  that  can  be  conveniently  studied  in  most 
curves  are  the  intercepts,  symmetry,  and  extent  of  the  curve. 

C 

^  21.    Intercepts.  —  The  intercepts  of  a  curve  are  the  dis- 

y  tances  from  the  origin  to  the  points  where  it  meets  the  axes. 
To  hnd  the  .T-intereepts,  set  y  =  0  and  solve  the  resulting 
equation ;  to  find  the  ^/-intercepts,  set  x  =  0  and  solve. 


y; 


ly     22.    Symmetry.  —  Definitions.     TJie  axis  of  symmetry  of 
^\/^tico  points  is  the  perpendicular  bisector  of  the  line  joining 
them.     The  center  of  symmetry  of  two  points  is  the  point  mid- 
way between  them. 

A  curve  is  symmetrical  with 
respect  to  an  axis  or  to  a  center 
ichen  each  point  of  the  curve 
has  its  symmetrical  point  on 
the  curve. 

Thus  the  points  (3,  4)  and 
(—3. 4)  are  symmetrical  with 
respect  to  the  2/-axis ;  (3,  4) 
and  (3,  —4)  with  respect  to 

the  a>axis ;  and  (3,  4)  and  (—3,  —  4)  with  respect  to  the 
origin.  From  the  figure  we  can  readily  establish  the  general 
principle : 

Two  2^oints  are  symmetHcal  ivith  respect  to  the  x-axis  ivhen 
their  abscissas  are  the  same  and  their  ordinates  differ  only  in 
sign  ;  to  the  y-axis  ichen  their  ordinates  are  the  same  and  their 
abscissas  differ  only  in  sign  ;  to  the  origin  ichen  their  resp)eG- 
five  coordinates  differ  only  in  sign. 

The  proof  is  left  to  the  student. 


Y 

^(-^vl/i) 

P.( 

^I'^i) 

^ 

L-— 

"x- 

P-i^-'^v-yi) 

A( 

^Wi) 

32 


ANALYTIC   GEOMETRY 


Applying  this  theorem  to 
curves,  we  see  that  a  curve 
is  symmetrical  w^ith  respect 
to  the  2/-axis,  if  for  each  point 
(x,  y)  on  the  curve,  the  point 
(—  X,  y)  is  also  on  the  curve. 
The  symmetry  of  a  curve 
may  be  determined  by  inspec- 
tion of  the  equation  accord- 
ing to  the  following  theorem  : 

If  the  substitution  in  an 
equation  of 

—  j:  for  X  ("gives  an  equation  reducible  to  itsl  i/-axis  ; 

—  y  for  y  \  original  form,  the  curve  is  sym-  r^-axis  ; 
—xioTX  and  —yiory  i metrical  with  respect  to  the       J  origin. 


The  proof  of  the  first  statement  is  as  follows : 

Let  the  equation  be  written  in  the  form  f(x,  y)  =  0.     Let 

P{Xi,  yi)  be  an}'  point  on  the  curve    and   Q(—  a:-,  2/1)  be  its 

symmetrical  point  with  respect  to  the  ^/-axis. 

By   hypothesis,    f(-x,  y)=f{x,  y)  ;    hence  f(-x„  y,) 

But  P  is  on  the  curve,  hence  /(x^,  y{)  =  0.  Therefore 
/(—  Xi,  2/1)  =  0,  and  hence  Q  is  on  the  curve. 

Thus  the  definition  of  symmetry  is  satisfied  and  the 
theorem  proved.  The  other  cases  are  treated  in  exactly 
the  same  manner. 

The  above  figure  is  the  graph  of  x'^—2y—l  =  0, 
an  equation  satisfying  the  test  for  symmetry 
with  respect  to  the  ^/-axis.  Solving  for  x,  we  have 
x=  ±  V2y  -h  1,  and  the  table  of  values  illus- 
trates various  pairs  of  symmetrical  points. 

A  curve  which  is  symmetrical  with  respect  to  both  the  x 
and  ?/-axes  is  symmetrical  with  respect  to  the  origin,  but  the 
converse  is  not  true  (e.g.  y  =  x^).     Moreover,  a  curve  may 


0 

±1  _ 
±  V3 

±  V5 
i3 


COORDINATES  AND   EQUATIONS 


33 


have  axes  of  symmetry  other  than  the  coordinate  axes.  In 
this  case,  if  there  are  two  perpendicular  axes  of  symmetry, 
their  intersection  will  be  a  center  of  symmetry. 

Exercise  10.  Prove  that  if  Pi(xu  yi)  and  Piixz,  2/2)  are  sym- 
metrical with  respect  to  the  x-axis,  then  x\  =  X2,  and  yi  =  —  y^. 

Exercise  11.     Write  a  proof  of  the  third  test  for  symmetry. 

2B.  Extent.  —  To  investigate  the  extent  of  a  curve,  we 
solve  its  equation  for  y  in  terms  of  x,  and  for  x  in  terms  of 
y.  If  either  operation  gives  rise  to  radicals  of  even  degree 
involving  one  of  the  variables,  say  x,  then  values  of  x  which 
make  the  expression  under  the  radical  negative  must  be  ex- 
cluded, for  the  corresponding  values  of  y  would  be  imaginary. 

To  illustrate,  consider  the  equation,  4  x^  —  9  ?/2  =  36. 

Solving,    y  =  ±  sVcc^  — 9,  x  =±  fvl/H-^. 

For  a;2  <  9^  or  for  vahies  of  x  between  ±  3,  x2_9  js  negative,  and  there- 
fore y  is  imaginary.  Hence  such  values  of  x  must  be  excluded  from  the 
table  of  values  and  the  curve  lies  wholly  without  that  part  of  the  plane 
bounded  by  the  lines  x—±3. 

As  2/2  4-  4,  >  0  for  all  values  of  y,  no  value  of  y  needs  to  be  excluded. 


PROBLEMS 


1.   Discuss  the  equation  4:x'^  +  y^  -  IQ  x  =  0  and  plot  its  graph. 
Solution.  — The  x-intercepts  are  0  and  4 


y 


0 
±3.4 
±4 
±3.4 

0 


Symmetry :  With  re- 
spect to  the  X-axis. 

X  =  2  ±  W16  —  ?/2. 

Hence  y^  must  be 
<  16  and  the  curve  lies 
between  the  lines-y  =  ±4. 


y 


=  ±  V4x(4-x). 


If  X  is  >  4  or  <  0,  the  factors 
under  the  radical  have  unlike  signs 
and  their  product  is  negative.  Hence 
such  values  of  x  must  be  excluded 
and  the  curve  lies  between  the  lines 
x  =  0  and  x  =  4. 


the  only  y-intercept  is  0. 

r 

7/ =4 


34 


ANALYTIC   GEOMETRY 


2.    Discuss  the  equation  y  =  x^  —  a^x  and  plot  its  graph. 

Solution.  —  This  equation  contains  an  arbitrary  constant  a.  It  will 
have  a  different  graph  for  each  value  given  a,  but  the  properties  obtained 
by  the  discussion  will  be  common  to  all  such  curves. 

Intercepts :  Putting  x  =  0,  the  y-intercept  is  0.  Putting  y  =  0,  x^  —  a^x 
=  0 ;  the  x-intercepts  are  0,  ±  a. 

Symmetry :  The  substitution  of  —  x  for  x,  or  —  y  for  y,  changes  the 
equation,  but  if  —  x  is  put  for  x  and  also  —  y  for  y,  we  get 

—  ?/=  (— x)3  — a2(  — x), 
which  reduces  to 

y  =  x^  —  a^x. 
Thus  the  curve  is  symmetrical  with 
respect  to  the  origin. 

Extent :  Since  this  equation  is  of 
the  third  degree,  it  will  have  at 
least  one  real  solution  for  any  value 
of  y.  Hence  we  shall  not  solve  it 
for  X.  It  is  already  solved  for  y  and 
no  radicals  appear,  hence  no  values 
of  X  need  to  be  excluded,  for  none 
make  y  imaginary.  Thus  x  may 
take  on  any  value.  If  the  equation 
is  put  in  the  form  y  =  x{x^  —  a^),  it 
becomes  evident  that  when  x  >  a, 
y  is  positive,  and  when  x  approaches 
infinity,  y  approaches  infinity. 
Thus  the  cui»ve  extends  to  infinity  in  the  first  quadrant.  Owing  to  the 
symmetry,  the  path  where  x  <  0  is  similar,  extending  to  infinity  in 
the  third  quadrant. 

We  could  now  sketch  the  curve  roughly.  A  table  of  values 
for  X  positive  gives  a  more  accurate  graph.  In  making  the 
table  we  assign  a  any  convenient  value,  as  1. 


X 

y 

0 

0 

1 

0 

2 

6 

3 

24 

etc. 

Discuss  each  of  the  following  equations  and  plot  their  respective 
graphs : 
.'  Z.   4x'^  —  y  =  0.  9. 

^■'      4.    2/^-8x+  16  =  0.  10. 


x2  -  ?/2  +  12  «/  =  0. 


j^.  x2  +  ?/2  -  16  a:  =  0. 

6.  a;2  +  4?/2  4-4?/  =  0. 

7.  x2  +  2y2  ^  16. 

>^.  9  ^2  _  a;2  _  36  =  0. 


11.  xy-9  =  0. 

12.  2  5C2/  +  15  =  0. 

13.  2/2  =  8x3. 
^4.  y  =  x^  —  9x. 


COORDINATES  AND   EQUATIONS  35 


15. 

X^  =  9  2/3. 

16. 

y  +  {x-  2)3  =  0. 

17. 

xy^  =  36. 

18. 

x(y'  +  1)  _  12  =  0. 

19. 

x^y  =  1. 

20. 

4  a:2  +  3  y2  =  0. 

21. 

y^  =  ax?. 

22. 

y-  =  2px. 

23. 

X2  +  2/2  =  (j2 

24. 

a2     62 

25. 

5c2     y2  _  ^ 
a^     62       • 

26. 

x^  =  2py. 

27. 

y  =  ax'\ 

28. 

xy  =  2  «2. 

24.  Symmetrical  Transformations.  —  In  case  the  tests  for 
symmetry  of  §  22  are  not  satislied,  the  substitutions  used  will 
transform  the  given  equation  into  a  new  equation. 

For  example,  if  we  substitute  —  x  for  x  in  the  equation  of 
a  curve  symmetrical  with  respect  to  the  ?/-axis,  the  equation 
is  unchanged ;  but  if  the  curve  is  not  symmetrical  with  re- 
spect to  the  y-Sixis,  a  new  equation  is  obtained.  The  ques- 
tion at  once  arises  :  what  relation  has  the  locus  of  the  new 
equation  to  that  of  the  given  equation '? 

In  this  case  it  can  be  shown  as  in  the  proof  of  §  22,  that 
for  any  point  P(xi,  ?/,)  on  the  locus  of  the  given  equation 
/(^'>  y)  =  ^)  the  symmetrical  point  Q(—  Xi,  y^  lies  on  the  locus 
of  the  new  equation  f(— x,  y)  =  0.  For  this  reason  the 
curves  are  said  to  be  symmetrical  to  each  other  with  respect 
to  the  y-Sixis,  and  the  transformation  is  called  a  symmetrical 
transformation.  Similar  reasoning  applies  to  the  other  sub- 
stitutions. 

Summarizing,  we  have  the  following  definition  and 
theorem : 

Two  curves  are  symmetrical  to  each  other  icith  respect  to  an 
axis  or  to  a  center,  ichen  each  point  of  one  curve  has  its  symmet- 
rical point  on  the  other. 

If  in  an  equation  substitution  is  made  of 

—  xioxx  { the  locus  of  the  new  equation )    y-axis  ; 

—1/  for  y    is  symmetrical  to  that  of  the  •    .r-axis ; 

—  JT  for  ^  and  —y  for  y  ^  old  with  respect  to  the  J     origin. 


36 


ANALYTIC    GEOMETRY 


Example  — Write  the  equation  of  the  curve  symmetrical  to 
y  =  x^  —  5x  +  Q  with  respect  to  the  ?/-axis,  and  plot  both  curves  on 
the  same  axes. 

Putting  —  X  for  x,  we  get  y  =  x^  +  5  x  -\-  6.  The  continuous  line 
is  the  graph  of  the  original  equation,  the  dotted  line  that  of  the  new 
one. 

Note  that  the  table  of  values  for  y  =  x^+5x  +  6  may  be  obtained 
from  that  for  y  =x^  —  6x  -\-  6  by  merely  changing  the  signs  of  the 
values  of  x. 


Tables  of  Values 

y  =  x^  —  5x  -{-  Q     y  =  x^  -\-  5x  -\-  6 


X 

y 

6 

X 

y 

0" 

0 

6 

1 

2 

-  1 

2 

2 

0 

-2 

0 

3 

0 

-3 

0 

4 

2 

-4 

2 

5 

6 

-  5 

6 

25.   Transformations  of  Equations  not  Altering  Loci.  —  The 

following  transformations  of  the  equation  leave  the  form  of 
the  locus  unchanged. 

(«)  Eearrangement  of  terms,  solving   for   one   variable, 
multiplying  by  a  constant. 

These  do  not  alter  the  relation  between  x  and  y. 

(b)  The  symmetrical  substitutions. 

(c)  Interchange  of  variables. 

The  effect  of  this  is  to  interchange  the  axes. 


r 


PROBLEMS 


1.    Write  the  equation  of  the  curve  symmetrical  to  the  given  curve 
with  respect  to  the  aj-axis,  and  plot  both  curves  on  the  same  axes. 

(a)  1/  =  3 x2  —  4  ;  (c)  y  =  x^  +  4:x; 

^b)  a;2  +  4  ^2  -  4  ^  =  0;  (d)  y  =  x^  -  2  x\ 

\y^2.   Write  the  equation  of  the  curve  symmetrical  to  the  given  curve 
with  respect  to  the  2/-axis  and  plot  both  curves  on  the  same  axes. 

i-fti)  ?/2  =  3x-4;  (c)   y  =  a;3  +  4a;; 

(6)  x2  +  4 1/2  -  4  X  =  0 ;  (d)  2/  =:  x3  -  2  x2. 


COORDINATES  AND   EQUATIONS  37 

^/^.    Write  the  equation  of  the  curve  symmetrical  to  the  given  curve 
with  respect  to  the  origin  and  plot  both  curves  on  the  same  axes. 

(a)  y  =  3a;2-2x  +  4;  ^    (c)  y  =  a:^  -  4x2; 

(&)  x  =  2^2_62/  +  8;  (d)  y  =  5x-3. 

4.    Plot  on  the  same  axes  the  graphs  of 

(a)  2/"^  =  3  X  +  5    and  x^  =  3  ?/  +  5  ; 
(&)    2/  =  (x  — l)3and  x=(y— 1)3; 
(c)  x2  —  4  2/2  =  4  and  'ip-  —  \x^  —\\ 
(rZ)  y2  =  ^3  and  x2  =  ?/3  ; 

(e)    y  =  7?  and  ?/  =  Vx. 


(-N ,  ■  26.  Equations  whose  Graphs  cannot  be  Plotted.  —  Since  real 
numbers  alone  can  be  used  as  the  coordinates  of  a  point,  an 
equation  which  is  satisfied  only  by  imaginary  values  of  the 
variables  does  not  define  a  curve  and  has  no  locus.  Such 
equations  may  be  most  easily  distinguished  by  completing 
squares.  If  then  every  term  is  positive,  the  equation  is  ob- 
viously satisfied  only  by  imaginary  values  of  x  and  y. 

PROBLEMS 

1.   Show  that  x2  -  2  X  +  2  2/2  +  8  ?/  +  14  =  0  has  no  locus. 

Solution.  — Completing  the  squares  in  z  and  y,  we  have 
(x-l)2  +  2  (2/ +  2)2  =  -5. 
For  any  real  values  of  x  and  y,  (x  — 1)^  and  {y  +  2)^  are  positive  or  zero-; 
hence  no  real  values  of  x  and  y  can  satisfy  the  equation. 

>-  2.   Examine  the  following  equations  for  the  existence  of  a  locus  : 
(a)   x2  +  6 X  +  2/2  _  4  ^  _^  17  _  0; 

/^A/)  y(^)  x2-3x  +  2/2  +  42,=_7; 

(c)  (x  + 2)2  4- j/2_6y  + 8=0; 

^^^  y(d)  x2  +  16  =  0; 

VV--J/A:  -(g)  a;4  +  4?/2  +  4  =  0; 

*^  ^     ,  (/)    X2  +  4X2/  +  42/2  +  16  =  0. 

3.  For  what  values  of  k  has  the  equation  x^  -\-  y'^  ■\-  k  =  0  no  locus  ? 

4.  What  is  the  locus  of 

(a)  x2  + 2/2  =  0; 

ih)  (X- 3)2 +(2/ +  3)2=0? 


38 


.  ANALYTIC   GEOMETRY 


5.  For  what  values  of  k  will  the  locus  be  imaginary,  a  point,  or 
a  curve  in  each  of  the  following : 

(a)  x2  +  92/2  =  ^•2-4; 

(6)  x'^-^x  +  y^  +  Sy-\-k  =  0; 

(c)  x'^  +  4:x  +  y'^  —  Q  y  -  k  =  0 ; 

(d)  x'^-9y^  +  k  =  0. 

27.  Intersection  of  Curves.  —  It  follows  from  the  definition 
of  the  locus  of  an  equation  that  a  point  lies  on  two  curves 
if  and  only  if  its  coordinates  satisfy  the  equation  of  each. 
Hence  we  conclude  that  the  coordinates  of  points  of  inter- 
section may  be  obtained  by  solving  the  equations  simul- 
taneously. If  there  are  no  real  solutions,  the  curves  do  not 
intersect. 

PROBLEMS 

1.  Find  the  points  of  intersection 
of     the     curves     x^  +  y-  =  16     and 

y'^  =  6  X. 

Solution.  —  Solving  simultaneously, 
a;  =  2  or -8.  When  a;  =  2,  y  =±^12; 
when  cc=  — 8,  ?/=±V— 48.  Thus 
there  are  but  two  pairs  of  real  solu- 
tions, giving  the  two  points  (2,  Vl2) 
and  (2,  —  Vi2).  The  figure  shows  the 
curves  and  their  points  of  intersec- 
tion. 

Find  the  points  common  to  the  loci  of  the  following  equations  and 
plot  the  loci. 

Note.  —  In  many  cases  the  points  of  intersection  and  the  intercepts, 
together  with  considerations  of  symmetry,  will  be  sufficient  for  plotting 
the  curves  without  calculating  tables  of  values. 

•  6.    a;2  +  2/2  _  25  =  0,   ^ 


2.   4x-y  +  2  =  0, 
X  -\-Sy  +  1Q  =  0. 

--/ .  3.   x-y  =  0, 

2/2  _|.  6  X  =  0. 

4.  2x-\-ny-15  =  0, 
7x-  by -12  =  0. 

5.  x  +  y  —  S  =  0, 
x^-^y  -\-2  =  0. 


y  -2x-\-n  =  0. 

7.    a:2  ^_  2/2  _  25  =  0, 
a;2  _  3  y  _  21  =  0. 

S.  8.    a:2  +  ?/  =  7, 
y2  _  81  =  0. 

9.    9x^  -1-4  2/2  =  36, 
9x^+lQy  =  33. 


COORDINATES  AND   EQUATIONS  39 

10.  a;2  _  y2  +  36  =  0,  14.   3a;2  +  4  2/2  :=  12, 
y-2-3x-3'2  =  0.  x^-y^  =  - 10. 

11.  y'^  =  2ax,  15.    y  =  x^, 

7?  =  ay.  y  =  x'^  -\-2x. 

12.  X-  +  6  ?/2  z=  20,  16.    x'^  +  y-  -Qx  =  0, 
xy  =  4.  y-  =  x^. 

13.  a:2  +  ?/2  -  8  a:  =  0,  17.   xy  =  4, 

2/2  _  8  X  -  2  =  0.  2/2  +  a;2  =  9. 


r7-=^ 


^ 


P) 


^h-^jJU^^r^, 


CHAPTER   II 

THE   STRAIGHT  LINE 

28.  Equations  of  the  Straight  Line.  —  Several  properties 
of  the  straight  line,  such  as  its  determination  either  by 
two  points  or  by  a  point  and  a  direction,  lead  to  relations 
between  the  coordinates  of  its  tracing  point  that  can 
be  expressed  in  the  form  of  an  equation.  There  are 
several  forms  of  the  straight  line  equation,  but  for  a  par- 
ticular line^  each  one  may  in  general  be  reduced  to  any  of 
the  others. 


29.   The  Point  Slope  and  Two-Point  Forms.  —  Let  the  line 
be  fixed  by  a  point  and  a  direction.     Let  the  point  be  P^{x,,  y^ 

and  let  the  direction  be  given 
by  the  slope  m  =  tan  a.  Let 
the  tracing  point  be  P{x,  y). 
By  the  slope  formula, 

m  =  ^  ~  ^S 

£C  -^  X^ 

or-   y-yi  =  m{x-Xi).      (7) 

This  is  the  point  slope  form 
of  the  straight  line  equation. 
It  is  used  in  writing  the  equation  of  a  line  when  one  point 
and  the  slope  are  known.  It  may  also  be  used  when  two 
points  are  known,  for  then  the  slope  of  the  line  can  be 
found  at  once  by  the  slope  formula  (3).  In  the  latter  case 
it  is  often  convenient  to  use  the  two-i)oiyit  form,  which  is  de- 
rived as  follows  ;     Let  the  line  be  determined  by  the  points 

40 


THE   STRAIGHT   LINE 


41 


Pi(xi,  yi)  and  P2(^2>  2/2))  and  let  the  tracing  point  be  P{x,  y). 
The  slope  formula  gives  us  at  once 

m  =  y  ~y^  and  m  =  ^ — ^  • 

Equating  these  we  have  the  desired  form, 


(7a) 


30.  The  Slope  Intercept  Form.  —  If  the  slope  and  the 
^/-intercept  are  given,  the  line  is  determined  by  a  point  on 
the  2/-axis  (0,  6),  and  the  slope  m.     Substituting  in  (7),  we 

have 


or 


•^  "}■. 


y  -b  =  m(x -  0),  J 

^y    y=mx^h,     //v^V-^sf 
which  is  called  the  slope  intercept  form. 

This  form  is  used  in  finding  the  equation  of  a  straight 
line  when  the  slope  and  the  ^/-intercept  are  known.  Con- 
versely, if  an  equation  can  be  reduced  to  form  (8),  its  locus 
is  a  line  whose  slope  is  the  coefficient  of  x  and  whose  y-m- 
tercept  is  the  constant  term  in  the  reduced  form. 

31.  The  Intercept  Form.  —  Suppose  the  intercepts  of  the 
line  are  given  ;  let  the  ^/-intercept  be  h  and  the  ic-intercept  a. 

Here  the  line  is  determined  by  two  points  (a,  0)  and 
(0,  h).     We  have  at  once, 


P{x,y) 


'111  =  — 


Then  by  (7) 


a 


which  reduces  to 

?  +  «=!. 
a     b 


(9) 


This  is  called  the  intercej^t  form.     It  is  used  in  writing  the 
equation  of  a  straight  line  when  the  intercepts  are  known. 


(  MAm-Q 


\ 


^ 


42 


ANALYTIC   GEOMETRY 


32.  Lines  Parallel  to  the  Axes.  —  Th^  eqaation  of  a  line 
parallel  to  the  ?/-axis  cannot  be  written  in  forms  (8)  or  (9), 
since  there  is  no  ^/-intercept.  The  equation  of  such  a  line 
is  obviously 


^    X  =  a. 


Similarly  the  equation  of  a  line  parallel  to  the  .T-axis  has 
the  form        /-'  , 

Exercise  1.     Derive  the  intercept  form  geometrically  *  when 

(a)  a  is  positive  and  h  negative  ; 
(6)  a  is  negative  and  h  positive  ; 
(c)  both  a  and  6  are  negative. 

Exercise  2.     Derive  the  slope  intercept  form  geometrically  *  for  the 
cases  given  in  Exercise  1. 

PROBLEMS 

1.    Derive  the  equation  of  the  line  determined  by  the  points  (3,  1) 
and  (5,  4).     Also  reduce  the  result  to  the  slope  intercept  form. 

Solution.  — By  the   slope   formula 
rn  =  f .    Using  the  point  slope  form, 

y-l  =  i(.T-3), 
which  reduces  to 
3x  —  2y  —  7  =  0. 

To  reduce  this  to  the  slope  inter- 
cept form,  solve  for  y : 

y  =  ^^  —  h 

This  form  shows  that  the  slope  is  f 
and  the  ?/-intercept  is  —  §. 

Equations  can  be  readily  checked 
by    substituting   the    coordinates   of 

known  points   or   drawing   the  line  from  the  intercepts  and  observing 

whether  or  not  it  satisfies  the  given  conditions. 

J^     2.    Derive  the  equation  of  the  line  determined  by  the  given  points 
in  each  of  the  following,  and  find  its  slope  and  intercepts. 

(a)   (-2,2),  (6,5);  .    (c)   (7,  o),  (_4,  -3); 

(&)   (4,  2),  (6,  -  5)  ;  ^.(d)   (8,  -  5),  (-  4,  3). 


*  I.e.  without  using  the  slope  formula. 


(0 

a  = 

-3, 

a 

~6  ' 

a  = 

-4, 

a 

27r 
~  3    ■ 

V(k) 

b  = 

-5, 

a 

=  tt; 

(0 

b  = 

6,  a 

= 

37r 
4 

^■"^  THE   STRAIGHT   LINE  43 

^Z.    Derive  the  equation  of  the  line 

(«)   Passing  through  (3,  5)  and  of  slope  | ; 
^^b)  Passing  through  (—  5,  2)  and  of  slope  —  i ; 

(c)  Passing  through  (—5,  —  1)  and  of  slope  2 ; 

(d)  Passing  through  (0,  0)  and  of  slope  —  | ; 

(e)  Passing  through  (5,  1)  and  parallel  to  the  y-axis; 
/(/)  Passing  through  (1,  5)  and  parallel  to  the  x-axis  ; 

(g)  Passing  through  (8,  2)  and  of  inclination  30°  ; 
(h)  Passing  through  (—  2,  5)  and  of  inclination  135°  ; 
(i)  Passing  through  (1,  5)  and  of  inclination  90°. 

Alls,  (a)  3x-2?/  +  1  =0;   (g)  x-  ^/Sy  =  8-2^/S. 

w  4.    Derive  the  equations  of  the  lines  from  the  given  data. 
.    ^  v     (a)  Intercepts  2  and- '— 6 ; 
/w^"""^^       lA^)  Intercepts  -  6  and  —  |  ; 

(c)  Intercepts  |  and  ^ ; 

(d)  Intercepts  —  |  and  2  ; 

(e)  m  =  I,  b=—6; 
/if)  m=-2,b  =  S; 
y  (g)   m  =  S,  a  =  4:  ; 

(^)  m  =—  2,  a  =  —  3; 

Ans.  (a)  6x- 2y. -12=0;  (i)  x-VSy -\-S  =  0.      • 

\^'r^.   Find  the  angle  between  the 'line*  2x  —  Zy  =  b  and  each  of  the 
lines  in  Problem  4.  ^  >    Uj^  ^««-    («)  37°  52'. 

.<  6.  Keduce  each  of  the  following  straight  line  equations  to  the  inter- 
cept and  slope  intercept  forms  if  possible,  and  draw  the  Unes. 

(a)  3x-4i/-8  =  0;  (e)4x-3y  =  0; 

..^N  ^_l=i.  (/)5x  +  |2/  =  4; 

•  /     '      ^^  ^  3      5         '  {g)  2a;-3?/  +  3  =  0; 

VW^    <(c)  Zx-2y  =  Q;  {h)x-Sy  =  0. 
^           )^{d)Sy +  2=0; 

-/s  7.  Find  the  equation  of  the  line  passing  through  the  point  (1,  3) 
and  forming  with  the  axes  a  triangle  of  area  8. 

8.    Show  that  all  lines  for  which  b  =  a  have  the  same  slope. 

<  9.  The  distance  between  two  fixed  points  is  2  c.  Find  the  equa- 
tion of  the  locus  of  a  point  moving  so  that  its  distances  from  the  fixed 
points  are  equal. 

*  More  properly  we  should  say  "  the  line  whose  equation  is  2  a;  —  3  ?/  =  5," 
but  the  close  relation  between  a  curve  and  its  equation  has  given  rise  to  the 
custom  of  using  the  abbreviated  phrase. 


44  ANALYTIC   GEOMETRY 

10.    The  distance  between  two  fixed  points  is  2  c.     Find  tiie  equa- 
tion of  tlie  locus  of  a  point  moving  so  that  the  difference  of  the  squares 
of  its  distances  from  the  fixed  points  is  k. 
j   ^         11.    The  base  of  a  triangle  is  fixed  in  length  and  position.     Find  the 
M      locus  of  the  opposite  vertex  if  the  slope  of  one  side  is  twice  the  slope 
of  the  other  and       (^d)  the  base  lies  on  the  a;-axis ; 
(b)  the  base  lies  on  the  ?/-axis. 

12.  Show  analytically  that  the  centers  of  circles  passing  through 
two  fixed  points  lie  on  a  straight  line. 

13.  Derive  the  straight  line  equation  x  =  7iy  -\-  a,  where  n  =  cot  a, 
and  a  is  thq^jintercept. 

J/a/-^  ,    xi.    Find  the  equations  of  the  two  lines  through  the  origin  which  tri- 
A  ^  sebt  the  triangle  formed  by  the  line  -  -f-  ^  =  1  and  the  coordinate  axes. 

'^  15.    Two  straight  lines  of  slope  —  1  are  tangent  to  a  circle  whose 

center  is  the  origin  and  whose  radius  is  4.     Find  the  equations  of  the 
lines. 

16.  Find  the  equation  of  a  straight  line  through  the  point  (4,  3) 
and  having  equal  intercepts. 

33.  The  Linear  Equation.  —  Theorem  I.  Every  straight 
line  is  defined  by  an  equation  of  the  first  degree,  in  one  or  two 
variables. 

Proof.  It  has  been  shown  that  every  straight  line  cut- 
ting the  ?/-axis  may  be  defined  by  the  equation  y  =  mx  -\-  b. 
In  case  the  line  is  parallel  to  the  2/-axis,  its  equation  is  of 
the  form  x  =  a.     Both  of  these  are  of  the  first  degree. 

Theorem  II.  Conversely,  the  locus  of  every  first  degree 
equation  in  one  or  tico  variables  is  a  straight  lirie. 

Proof.  The  general  form  of  the  equation  of  the  first 
degree  is  ^x  +  5y  +  C  =  0,  (10) 

where  A,  B,  and  C  may  have  any  values,  zero  included. 

When  B  =^  0,  this  may  be  solved  for  y,  giving 

^  B^      b' 


THE   STRAIGHT   LINE  45 

This  is  of  the  form  y  =  mx  +  h  and  therefore  defines  a 

—  4  —  C 

straight  line  with  slope  —^  and  t^-intercept  -— -  • 

C 
When  B  =  0,  vre  have         x  = • 

This  is  of  the  form  x  =  a  and  therefore  defines  a  straight 

—  C 
line  parallel  to  the  ^/-axis  and  with  ic-intercept  — — .     Thus 

the  theorem  is  proved  for  both  cases. 

Note.  —  For  this  reason  an  equation  of  the  first  degree  is  called  a 
linear  equation,  the  term  being  used  by  analogy  for  all  equations  of  the 
first  degree  in  any  number  of  variables. 

Exercise  3.  Transform  the  general  equation  Ax  -\-  By  -\-  C  =  0 
into  the  intercept  form.     When  is  this  impossible  ? 

34.  Relations  between  Linear  Equations.  —  Theorem  I.  If 
ill  two  linear  equations  in  two  variables  the  coefficients  of  the 
variables  are  proportional,  the  lines  defined  by  those  equations 
are  parallel,  and  conversely. 

r,  W       -    "' 

Proof.     Let  the  equations  be  ^  y  ^    / 1,   ^ 

Ax  +  57/  +  O  =  0, 
and  A'x  +  B'y  +  C  =  0. 

Solving  for  y,  I  V 

AC.  A        C 

11  = X and  y  =  —  -^,x  —  ^;r,' 

^  B        B  B'        B' 

These  are  of  the  form  y  =  mx  +  b. 

Hence  m  =  —  —       and      m'  =  —  —  • 

But  by  hypothesis  A:  A  =  B  :B'  oy  A:  B  =  A  \  B'. 

Thus  we  have  ??i  =  m',  and  the  lines  having  equal  slopes 
are  parallel.       A/t  -   / 3  G)   -  ^ 

The  converse  and  the  special  case  where  B  =  B*  =  0  are 
left  to  the  student. 


Exercise  4.     Assume  the  lines  parallel  and  prove  that  the  coeffi- 
cients of  X  and  y  are  proportional. 

^  V-  <^H  -f  I V-  ^     ^^  -  i  %^  ^ 


46  ANALYTIC   GEOMETRY 

Theorem  II.     In  any  tico  linear  equations, 
Ax  +  By+  (7=0, 
A'x  4-  B'y  +C'  =  0, 
if  AA'  =  —  BB',  the  lines  defined  by  the  equations  are  x>erpen- 
dicular,  and  conyerselv. 

Proof.     Solving   the    given   linear    equations    for   y,   as 

before,  we  have 

A  ,      ,  A' 

m  = ana  m  = • 

B  B' 

m.                          ,     f     A\r     A'\      A  A' 
Then  ?)i?7i  = 


BJ\     B'J      BB' 

But  by  hypothesis  AA'  =  —  BB',  hence  mm'  =  —  1,  and 
the  condition  of  perpendicularity  is  satisfied. 

Exercise  5.     Assume  the  hnes  perpendicular  and  prove  that 

AA'  =  -  BB'. 

Theorem  III.     If  in  any  tivo  linear  equations 
Ax  +  By  +  C=0, 
A'x  +  B'y  +  C  =0, 
we   have   A:A'  =  B:B'=C:C',    the   lines   defined    by   the 
equations  are  identical,  and  conversely. 

Proof.     Prom  the  assumed  proportions 

^  =  9L   ovb  =  b\ 
■     B     B'' 

Also  A     A'  ,       T    f 

—  =  — - ,  or  m  =  ??i  as  beiore. 

B      B" 

Hence  the  lines  have  the  point  (0,  b)  in  common  and  also 
have  the  same  direction.     They  are  therefore  identical. 

Exercise  6.     Assume  the  lines  identical  and  prove  that 
A:A'  =  B:B'=C.C'. 

With  the  aid  of  these  theorems  it  is  possible  to  determine 
by  inspection  of  their  equations  whether  two  lines  are  iden- 


/ 


THE   STRAIGHT   LINE  47 


tical,  parallel,  perpendicular,  or  otherwise.  This  is  of  con- 
siderable aid  in  checking  the  solutions  of  problems  involving 
straight  lines.  Suppose,  for  example,  that  it  is  desired  to 
write  the  equation  of  a  line  passing  through  the  point 
(4,  —  2)  and  perpendicular  to  the  line  10  x  —  4  ?/  +  11  =  0. 

From  the  given  equation  the  slope  of  the  line  is  f,  and 
hence  the  slope  of  the  required  line  is  —  |  (§  12).  Substi- 
tuting in  the  point  slope  equation  and  reducing,  we  get 
2x  +  oy-{-2=0.  This  line  is  evidently  perpendicular  to 
the  given  line,  for  AA' =  20  and  BB'  =  —  20.  Also  it  passes 
through  the  given  point,  for  2«4-}-5-— 2  +  2  =  0. 

The  same  problem  may  also  be  solved  by  using  Theorem 
II.  Any  line  having  the  form  4:X-\-10y  =  k  will  be  per- 
pendicular to  the  given  line  10  x —  4.y -\- 11  =  0.  Since  the 
line  passes  through  (4,  —  2), 

A;  =  4  .  4  + 10  ( -  2)  =  16  -  20  =  -  4. 
Hence  the  required  equation  is 

4:X-\-10y  =  —  4:,ov2x-\-oy-{-2  =  0. 

(\ 

35.   Application  to  the  Solution  of  Linear  Equations.  —  Two 

equations  of  the  tirst  degree  in  two  variables  are  classified  as 
(a)  Simultaneous   if   they   have    one    solution,    w^hich   is 

usually  the  case ; 

(6)  Incompatible  if  they  have  no  common  solution,  as 

'  3x  -\-7y  =  1, 
\6x  +  Uy  =  l', 

(c)  Dependent  if  they  have  innumerable  solutions,  as 

f  9aj—  6?/  =  3, 

[Zx-2y  =  l. 

It  is  sometimes  convenient  to  determine  the  number  of 
solutions  without  solving.  The  test  by  which  the  class  of 
the  equations  can  be  determined  is  easily  deduced  from  the  re- 
lation between  their  loci^  and  from  Theorems  I  and  III,  §  34. 


48  ANALYTIC   GEOMETRY 

(a)  The  equations  have  one  and  only  one  solution,  if  their 
lines  intersect.     This  is  the  case  when 

(b)  The  equations  have  no  common  solution,  if  their  lines 
are  parallel.     This  is  the  case  when 

A:A'  =  B:B'. 

(c)  The  equations  have  innumerable  solutions,  if  their 
lines  have  more  than  one  point  in  common  and  are  there- 
fore identical.     This  is  the  case  when  ■ 

A:A'  =  B:B'  =  C:C\  y-/-  2^- 

PROBLEMS 

ji^.   Select  two  pairs  each  of  parallel  and  perpendicular  lines  from 
wie  following : 

(a)  y  =  2x  +  ^;  (e)  x -\- 5y  =  0; 

(b)  2y  =  -6x-\-9;  (/)  6a:  +  2^  +  l  =  0; 

(c)  y  =  2«  +  9;  (g)  5x-y-S  =  0] 

{d)  2y=-\x-\-7  ;  <ih)  x-{  2y-l  =0. 

2.  Identify  the  following  pairs  of  equations  as  simultaneous,  in- 
compatible, or  dependent : 

(a)  2x+    6?/-7  =  0,      (6)2x  +  5?/  =  3,     (c)        x-Sy=8, 
6x  +  lSy  -\-8=0;  2x-by=S;  12y  -ix=-S2. 

3.  Show  that  the  equations  2  Ax  +  By  +  D  =  0  and  Bx  +  2  Cy  ■}• 
F  =  0  are  simultaneous  ii  B'^  —  4:  AC  ^  0. 

4.  Find  the  equations  defining  the  lines  determined  as  specified 
and  verify  the  result.     In  each  case  two  lines  are  to  be  found. 

v/.  (a)  Both  of  the  required  lines  pass  through  (3,  0)  and  are  respec- 
tively parallel  and  perpendicular  to6x  —  5!/  +  10  =  0; 

(b)  As  before,  using  (—2,  6)  and  8x  +  7y+12  =  0; 
.(c)  As  before,  using  (4,  3)  and  3  x  +  2  y  =  3  ; 

v;(d)  As  before,  using  (0,  0)  and  2x  —  5y  =Q  ; 
(e)  As  before,  using  (2,  0)  and  S  x  -{-  2  y  =  15  ; 
(/)  As  before,  using  (0,  0)  and  5x  -\-  by  =  12  ; 
(g)  As  before,  using  (3,  3)  and  Sx  —  2y  =  10 ; 
(h)  As  before,  using  (0,  0)  and  x  +Q  =  0. 

Ans.    (a)  6x-5y- 18  =  0;  bx -^  Gy -^15  =  0. 

i 
*  This  symbol  has  the  meaning,  "  is  unequal  to." 


\M.cA:>^'{jjA^A  V) 


THE   STRAIGHT   LINE  49 

5.    Show  that  the  bisectors  of  the  angles  of  any  rectangle  form  a 
square. 

^    6.   In  the  triangle  whose  vertices  gire  (0,  0),  (6,  0),  and  (8,  6)  find 
w^^  the  equations  of  its  sides ;      y^y\^    ^^-'^  5 
^**K^  the  equations  of  the  perpendicular  bisectors  of  the  sides  ; 

.  (c)  the  equations  of  the  perpendiculars  from  the  vertices  on  the 
opposite  sides ; 

(d)  the  equations  of  the  medians. 

^/V.    Show  that  in  the  above  problem 

p^a)  the  perpendicular  bisectors  meet  in  a  point ; 

(6)  the  medians  meet  in  a  point; 
,.{c)  the  perpendiculars  from  the  vertices  on  the  opposite  sides  meet 
in  a  point. 

(In  each  case  show  that  the  coordinates  of  the  intersection  of  two 
lines  satisfy  the  equation  of  the  third.) 

8.    Show  that  the  three  points  of  intersection  found  in  the  above 
problem  lie  in  a  straight  line. 

^^  9.    Show  that  the  medians  of  any  triangle  meet  in  a  point.     (Take 
"^he  vertices  as  (0,  0)  (a,  0),  and  (6,  c).) 

--•  10.    Show  that  the  perpendicular  bisectors  of  the  sides  of  a  triangle 

meet  in  a  point. 

-?>-: —  11-    Show  that  the  perpendiculars  from  the  vertices  of  a  triangle  on 
the  opposite  sides  meet  in  a  point. 

12.  Show  that  the  three  points  of  intersection  found  in  Problems 
9,  10,  and  11  lie  in  a  straight  lin6. 

13.  The  vertices  of  a  parallelogram  are  (0,  0),  (6,  0),  (4,  8),  and 
(10,  8).  Lines  are  dra^NTi  from  two  opposite  vertices  to  the  mid-points 
of  opposite  sides.  Show  that  they  are  parallel  and  that  they  trisect 
one  of  the  diagonals. 

14.  Find  the  center  and  radius  of  a  circle  circumscribed  about  the 
triangle  whose  vertices  are 

(«)   (0,0),  (-6,0),  (8,4); 
(6)   (4,0),  (-4,0),  (6,8). 

Ans.    (a)  center  (—3,  16),  radius  \/265. 

!6**-^15.    Two  opposite  vertices  of  a  square  are  (—  2,  3)  and  (4,  —  5). 
Find  the  other  vertices  and  the  equations  of  its  sides. 

^       16.    Find  the  fourth  vertex  of  the  parallelogram  which  has  three  'of 
its  vertices  : 

(a)   (3,  5),  (4,  9),  (-  2,  6)  ;  (6)   (8,  6),  (5,  -4),  (0,  3). 


50 


ANALYTIC   GEOMETRY 


17.    Find  the  equation  of  a  straight  line  through  the  origin  making 
an  angle  of  45°  with  each  of  the  lines  of  Problem  1. 

Hint.  — To  find  the  slope  of  the  required  line,  observe  that  its  inclina- 
tion is  45°  greater  or  less  than  that  of  the  given  line. 

Ans.    (a)  y=—  ^xov  y  =  \x. 


Y 

z 

•A 

B 

\r/ 

/ 

b 

V 

\ 

^ 

0 

a 

^\r 

36.  Geometric  Conditions  Determining  a  Line.  —  In  eacli  of 

the  standard  equations  y  =  mx  +  h  and  -  -}-  f  =  1,  we  note 

ah 

that  there  are  involved  two  arbitrary  constants  which  rep- 
resent geometric  conditions 
determining  the  line.  Two 
of  these,  the  intercepts  a 
and  5,  represent  distances, 
while  the  other,  the  slope 
771,  represents  a  direction. 
The  direction  may  also  he 
given  by  the  inclination  a. 
There  are  also  two  other 
such  constants  obtained  as 
follows.  Let  AB  be  the  given  line  and  OZ  a  line  from  the 
origin  perpendicular  to  it  at  C.  This  is  called  the  normal 
axis  of  the  line.  Its  inclination  is  denoted  by  w,  and  the 
perpendicular  distance  OC  of  the  line  from  the  origin  is  de- 
noted by  p  and  called  the  normal  intercept. 

Thus  we  associate  with  any  straight  line  six  constants ; 
three,  a,  6,  and  p,  representing  distances,  and  three,  m,  a, 
and  w,  representing  directions.  Any  pair  of  them  (one  being 
a  distance)  will  determine  the  line  (and  also  the  other  con- 
stants), and  combined  with  its  current  coordinates  they  will 
furnish  an  equation  of  the  line.  Of  the  many  possible 
equations  involving  these  constants  only  three  are  commonly 
used,  two  of  which  have  been  discussed  in  §§  30,  31,. 

37.  The  Normal  Form.  —  In  this  form  of  the  equation  the 
line  is  determined  by  x>  and  w. 


^  m    y-f^r^^ 


THE   STRAIGHT   LINE 


51 


To  derive  it,  take  any  point  P  on  the  line  and  draw  FD 
from  the  foot  of  the  ordinate  of  P  perpendicular  to  the 
normal  axis  at  D ;  also 
draw  PE  perpendicular  to 
FD.  It  follows  at  once 
that  Z  EFP  =  w, 

OD  ,^  l!}-p 


and 


cos  (i) 


sm  cu  = 


X 

EP 

y 


But 
OD+EP=x>. 


hy^-  .-j'-'^j 


Substituting,      a:  cos  w  +  i/  sin  (o  —  />  =  0. 

This  is  the  normal  equation.  Here  w,  the  inclination  of 
the  normal  axis,  like  a,  the  inclination  of  the  line,  is  reckoned 
from  0°  to  180°.  The  upward  direction  of  p,  reckoned  from 
the  origin,  is  taken  as  positive,  and  the  downward  direction, 
reckoned  from  the  origin,  is  taken  as  negative. 

38.   Reduction  of  a  Linear  Equation  to  the  Normal  Form.  — 

Let  the  equation  of  the  given  line  I  be 

Ax^-By-^C=  0. 
It  is  required  to  find  the  equation  of  I  in  the  normal  form. 

A 

Since  the  slope  of  the  given  line  is ,  the  slope  of  the 


B 


B 


normal  axis  is  the  negative  reciprocal  of  this,  — 

A 

tan  o)  =  —  •     Elementary  trigonometry  gives  us  at  once 
A 


Hence 


sina;  = 


B 


cos  CO  = 


A 


±V^2  +  ^'  ±^A'-\-B' 

Since  w  <  180°,  sin  w  is  positive.  Therefore  the  sign  of  the 
radical  must  be  taken  to  agree  with  that  of  B.  Thus  we 
have  determined  sin  w  and  cos  co  in  terms  of  the  coefficients 
of  the  given  equations. 


52 


ANALYTIC   GEOMETRY 


Dividing  the  given  equation  by  V^^  +  ^'^  we  liave 
A  .  B  .  C 


-X 


:2/  + 


=  0, 


which  is  of  the  form 

X  cos  (0  +  2/  sin  w  —  j>  =  0. 
Comparing  these  equations,  we  see  that  p  must  be 

-O 


■VA'  +  B' 
The  special  cases  where  A  =  0  and  ^  =  0  are  left  to  the 
student. 

Summing  up,  we  have  the  following  working  rule : 

To   reduce   the   linear    equation    Ax -\-  By  -{- C  =  0   to   the 
normal  form,  divide  it  by   ±  VA^  +  B^,  giving  the  radical  the 

sign  of  B. 

Example.  —Reduce  3x— 4  «/=20 
to  the  normal  form. 

Here  VA^  +  B^  =  5,  and  B  is 
negative.  Therefore  the  required 
equation  is 

o         5 

This  gives  cos  w  =  —  f ,  sin  w  =  f , 
and  p  =  —  4.  Plotting  the  equa- 
tion affords  an  excellent  check  on 
the  work. 

The  equation  Ax-\-By+C=0  can  always  be  reduced  to 
the  normal  form.  For  the  division  by  the  radical  is  always 
possible,  since  A  and  B  cannot  both  be  zero.  This  consti- 
tutes one  of  the  two  most  important  advantages  of  the 
normal  form.  The  other  consists  in  its  convenience  for  use 
in  finding  the  distance  from  a  line  to  a  point. 

Exercise  7.     Derive   the  normal  equation  for  a  line  crossing  the 

axes 

(a)  in  the  second  quadrant ;  (c)  in  the  fourth  quadrant. 

(6)  in  the  third  quadrant; 


THE   STRAIGHT   LINE  53 

PROBLEMS 

)( 1.   Draw  the  lines  satisfying  the  following  conditions  : 
^£^   (a)  a  =  -5,  a=    30'^;  (/)  i>  =      3,  a,  =    45°; 

^^n(&)   6=       4,«=    60°;  ig)    a=-6,p  =  S; 

\r^  \c)  p  =  -  3,  «  =  120° ;  (/i)    6  =  -  5,  p  =  -  4 ; 

(cZ)  a  =       5,  w  =  150° ;  (i)     «  =  -  3,  ;>  =  -  3. 

(e)  6  =  -  4,  w  =  120°  ; 

3.    With  ri^Zer  and  compasses  construct  the  required  line,  given 


(a)  a  and  «  ;  {d)    a  and  w 

(&)  h  and  «  ;  (e)    6  and  w 

(c)  j9  and  «;  (/)  p  and  w 


(^)   a  and  p  ; 
(Ji)  h  and  p. 


^3.   Write  the  equations  of  the  lines  of  Problem  1. 

(^US^  T      Ans./{a)  x-V3^  +  5  =  0;      {g)  3cc-4?/  +  15  =  0. 

'>^4.    Find  the  distance  of  a  line  from  the  origin  when  the  intercepts 
are : 
\  ^(a)  3  and  4  ; 

(6)    3  and  -  4  ; 

(c)    —  6  and  8  ; 

y.5.   Find  the  equations  of  the  lines  from  the  following  data  :         ^, 
>((a)  p  =  -  6,  w  =  45°  or  135°  ;  ,    '    .       ^  /I 

(&)  J9=       5,  w  =  150°;  ^   '  ' 


(d)    6  and  9 ; 

(g)  7  and  2  ; 

■X(e)    5  and  —  12  ; 

(h)  7  and  Vl5. 

(/)   -5  and  12; 

A)is.   (a)   V 

(c)  ;9  =      4,  «  =  60°  or  120° ; 
{d)  p=      3,  w  =  0°,  90°,  or  180°; 

(e)   p=       6,  a>  =  75°.     (sin  75°  =  cos  15 


3o^Jl  +  cos30°  N 


1^6.    Write  the  equation  of  the  line  passing  through  the  point  (5,  — 
10)  and  having  ^ 

^a)   0,^135°;  (c)  p=  -2; 

^/)oih>  (&)    0,  =  120° ;  {d)  p=-6. 

Ans.   (a)  x-y  -15=0;     (c)  4  x  +3  ?/  +  10  =  0. 


y^7.   Reduce  to  the  normal  form  and  verify  the  work  by  constructing 
the  line  from  its  intercepts  in  each  of  the  following  cases : 
<a)     4x-   Sy=12;  ^^^  3.^V2  +  2/; 

^b)     8x+15,  =  12;  (^^c  +  V3,ilO; 

(c)     3x+   4y  =  12;  r%  6^^-8^  =  25 

(d)12.+    5,=    0;  (o\-  +  2/+5  =  i. 

(e)       X-      y=    0; 


54  ANALYTIC   GEOMETRY 

8.    Find  p  and  tan  w  in  terms  of  a  and  h. 


— ^^3;;;^;^^^ ;  tan  oj  =  —  • 


Ans.  p  = 

H^  V    9.    Prove  tan  w  =  —  cot  a. 

'■^  /  10.    Find  tan  w  in  terms  of  A  and  5. 

v/ll.   Derive  the  intercept  form  of  the  straight  line  equation  from  the 
normal  form. 

12.   Find  the  coordinates  of  the  foot  of  the  perpendicular  from  the 
origin  to  a  line  in  terms  of  its  intercepts  and  inclination. 

39.  Distance  from  a  Line  to  a  Point.  —  Let  I  be  a  line  of 
which  the  equation  is  given  and  P^  a  point  whose  coordi- 
nates (cci,  2/1)  ^^6  given,  and 
let  d  be  the  distance  NPi 
from  the  line  to  the  point, 
reckoned  from  the  line. 

Through  Pi  draw  the  line 
?!  parallel  to  I  and,  if  neces- 
sary, produce  the  normal  p 
to  meet  it.  The  normal 
inclination  tu  is  the  same 
for  I  and  l^,  and  is  known 
from  the  equation  of  I 
when  it  is  reduced  to  the  normal  form.  The  normal  in- 
tercept of  Zi  is  OF  and 

0F=  OE  4-  EF  =  2^  -h  d. 

Hence  the  normal  equation  of  l^  is 

p  -}-  d  =  X  cos  w  +  y  sin  w. 

This  is  satisfied  by  the  coordinates  of  P^ ;  hence 

p  -^  d  =  Xi  cos  (o  -h  2/1  sin  w,  or 

d  =  Xi  cos  0)  -f  i/i  sin  (n  —  p.  (12) 

But  the  equation  of  I  is 

X  cos  w  -f  2/  sin  0)  —  p  =  0.  (11) 


h' 


7 


THE    STRAIGHT   LINE 


55 


Thus  the  first  member  of  (11)  becomes  the  value  of  d 
when  the  coordinates  of  P^  are  substituted  for  x  and  y. 
Hence  follows  the  rule  : 

To  find  the  distance  from  a  line  to  a  point,  reduce  the  equa- 
tion of  the  line  to  the  normal  form  and  substitute  the  coordinates 
of  the  given  point  for  the  variables.  The  value  of  the  first 
member  is  the  required  distance. 

Example.  —  Find  the  distance  of     P{-1,6) 
the   point   P(—  1,  6)   from   the   line 
2x  — 3y=10. 

The  normal  equation  of  I  is 
10 


2^  I    3^  I 


Vl3 
'.  d  = 


(-2)(-l)   I  3 


=  0. 

6 


V13 


+ 


10 


Vl3      Vl3 


30 
v/l3 


Here^,  being  negative,  is  reckoned 
downward  from  the  origin  and  cZ,  being  positive,  is  reckoned  upward 

from  the  line  I. 

Exercise  8.     Show  that  the  distance  formula  can  be  written 
^^  ^  Axi  J^Byi+  C 
±  ^A^  +  B^ 
where  the  radical  should  have  the  sign  of  B. 


y. 


PROBLEMS 

'^1.    In  each  case  find  the  distance  from  the  line  to  the  point,  and 
verify  results  by  constructing  a  figure. 

(a)  3x-4?/=10,  (2,  1);  "^  {d)  3x  +  4t/  =  12,  (4,0);  , 

))  x  =  ?/,  (0,  4);  (e)  5a: +  12?/ =  5,  (0,0); 

[c)  4a:-2y  +  5  =  0,   (-3,  6);      (/)  x  +  y  +  1  =  0,  (4,4).^ 

V".    •' 

K2.   Find  the  distance  between  the  parallel  lines  >    r  •     •' 

{a)y  =  ?>x  +  b,  *^)3x  +  2y  =  2,       '    <c)  6x  +  9?/ =  0,    •';;/;  y 

y  =  Zx  —  b\  3x  +  2?/  =  6;  4x4-6?/=-^  8.'    '■,!>«', 

/l*1-  ^^s3  Ms.    (a)   VTO.-^'-'),' 


/ 


VA 


56 


ANALYTIC   GEOMETRY  y  f^^^h^-^^^ 


Find  the  distance  from  the  intersection  of  the  lines 
(a)     x-4y-  12  =  0 

3  X  —  4  ?/  -    2  =  0    to  the  line  a-  +  ?/  +  10  =  0  ; 

^(6)  Sx-    y-    2=0 

5x+     y—12  =  0    to  the  line     12  x  —  5?/ +  23  =  0. 

S^4.   Find  the  areas  of  the  triangles  formed  by  the  three  lines  in  (a) 
and  (6)  in  Problem  3.  Ans.    (a)  jf^. 

5.  Find  the  areas  of  the  triangles  whose  vertices  are 

(a)   (0,  0),   (1,  2),   (5,  7)  ;  (c)   (-  7,  8),   (-  7,  -  6),   (1,  -  4)  ; 

(6)   (-1,3),   (2,-4),   (0,6);     (d)   (5,4),  (8,3),  (6,5). 

Ans.    (a)  I  ;  (c)  56. 

6.  Find  correct  to  one  decimal  place  the  areas  of  the  polygons 
having  the  following  vertices  : 

(a)   (6,0),   (7,10),   (-3,5),   (-2,  -6); 

(6)   (2,  0),   (1,  4),   (-  2,  2),  (-  1,  -  3),    (1,  -  2). 

Ans.    (a)  94.5;  (6)  17.5. 

7.  Find  the  equations  of  the  bisectors  of  the  angles  between  the 
lines  Sx  —  4y  =  12  and  12x  +  5y  =  S0. 

Solution. — Let  the  lines  meet  at  E  ;   there  are  two  bisectors  EF  and 

EG.  Let  Piixi,  ?/i)  lie  upon 
EG,  the  bisector  of  the  angle 
BEC.  Then  by  plane  geometry 
the  perpendiculars  HP^  and 
KPi  are  of  equal  length. 

To  find  the  length  of  HPi 
put  3  X  —  4  ?/  =  12  into  the  nor- 
mal form. 


Then 
HP,= 

Similarly 


OXi 

5 


^Ml 


13 


13 


30 
13' 


But  HPi  is  negative  and  KPi 

positive. 

Therefore 

HPi^  =  -  KPi,  or 
12  Xi       5yi    .   30 
5  13  13        13* 

Simplifying  and  dropping  primes,  21  a;  +  77  ?/  +  6  =  0  is  the  equation  of  EG. 
In  like  manner  the  equation  of  EE  is 

99x  — 27?/ -306  =  0. 
A  convenient  check  on  the  accuracy  of  the  work  is  the  application  of  the 
test  for  perpendicularity  to  the  two  solutions. 


THE   STRAIGHT   LINE  57 

^^^"^y^  8.    Find  the  equations  of  the  bisectors  of  the  angles  between  ,       ^  ■ 

(a)  3a:  +  42/  =  5,  (6)  3x  -  4?/ ==  7,    ^  ..       "^  ^ 

4a;  +  32/  =  6;  4x-32/  =  8. 

*       x/^.   Find  the  locus  of  points  twice  as  far  numerically  from  the  first 
line  as  from  the  second  in  each  of  the  groups  in  Problem  8. 

Ansrjsia)  5x-\-27j^7  =  0,  llx -f  lOy  -  17  =  0.        /^^^ 

y  10.  Find  the  locus  of  points  the  sum  of  whose  distances  from  each 

of  the  pairs  of  lines  in  Problem  8  is  12. 

Ans.'{a)  7x  +  7y  —  71  =0. 

11.  Find  the  locus  of  points  6  units  distant  from  the  line 

(a)     3x  +  4?/  =  5;  (c)  15x  -  Sy  =  12. 

(6)  12x-  oy  =  n  ; 

Ans.    (a)  Sx-\-iy  =  S6,     Sx  +  4:y  =  -25. 

12.  A  triangle  has  vertices  (3,  0)  and  (—  1,  3).  Find  the  locus  of 
the  third  vertex  if  the  area  is  to  be  16. 

13.  Find  the  equation  of  the  line  halfway  l)etween  the  parallels 

{a)  2x  +  y  =  7,  (b)  Sx-2y  =  9, 

2x+y  =  10;  Sx-2y  =  l6. 

^    14.    A  triangle  has  the  vertices  (0,  0),  (12^,  0),  and  (8,  6). 
y  (a)  Find  the  equations  of  the  bisectors  of  the  angles. 
(6)  Show  that  these  bisectors  meet  in  a  point. 

(c)  Find  the  coordinates  of  the  center  and  the  radius  of  the  inscribed 
circle. 

40.  Systems  of  Straight  Lines. — When  one  of  the  con- 
stants of  a  straight  line  equation  is  arbitrary  and  the  other 
absolute,  as  in  y  =  3x  +  k,  the  equation  defines  a  line  for 
each  value  assigned  to  the  arbitrary  constant,  but  each  line 
so  defined  has  the  property  given  by  the  numerical  con- 
stant. 

Thus  in  the  above  equation  k  is  the  ^/-intercept,  and  it 
may  have  any  value ;  but  the  slope  of  each  line  is  3,  and  by 
assigning  all  possible  values  to  k  we  have  all  possible  lines 
of  slope  3.  Such  a  group  of  lines  is  called  a  system,  and  the 
arbitrary  constant  is  called  the  parameter  of  the  system. 

We  have  already  seen  that  any  of  the  standard  forms  of 
the  straight  line  equation  involves  two  arbitrary  constants 


58  ANALYTIC   GEOMETRY 

which  correspond  to  the  geometric  conditions  defining  the 
line.  Hence  if  one  of  these  constants  is  given  a  numerical 
value  and  the  other  left  arbitrary,  vre  get  a  system  of  straight 
lilies  characterized  by  the  geometrical  property  defined  by 
the  numerical  constant,  and  with  one  geometrical  property 
left  arbitrary  or  variable.  This  correspondence  is  general, 
as  will  be  seen  later  in  the  case  of  the  circle  and  the  conies. 

PROBLEMS 

In  each  of  the  following  problems  draw  several  lines  of  the  system. 

1.  Write  the  equations  of  the  systems  of  lines  defined  as  follows: 
(a)  passing  through  (5,  —  1)  ;  Ans.  ?/  -f  1  =  m{x  —  5). 
(&)  passing  through  (—  2,  2)  ;           (/)  having  the  slope  —  |; 

(c)  passing  through  (6,  —  3)  ;  {g)  having  the  slope  I ; 

(c?)  with  the  ^/-intercept  —  2  ;  {h)  distance  from  origin  13  ; 

(e)  with  the  a;-intercept  —  3  ;  (i)   distance  from  origin  5. 

2.  What  geometric  conditions  define  the  systems  of  lines  repre- 
sented by  the  following  equations  • 

Hint.  —  Put  each  equation  in  a  standard  form. 
^y    {a)  x  +  k  =  0;  {e)  2x-ky  +  2  =  0; 

(6)  3  :k  +  4  y  +  ^•  =  0  ;  lf)kx  +  ky  -Q  =  0  ; 

>  (c)  x  +  ?jky-b  =  0;  ^f  gr)  kx  +  !/ VH^  =  6 ; 

{d)  kx-\-2y  -Q  =  0;  (^h)  y  -  1  =  mx  +  2 m. 

Ans.     (c)  a:-intercept  5 ;     (g)  normal  6. 

«  3.   Determine  k  so  that 

1  (a)  the  line  Sx  +  iy  =k  passes  through  (—2,  1)  ; 
(6)  the  line  Sx  —  4y  =  k  passes  through  (—  3,  4)  ; 
'      (c)   the  line  kx  —  4y  =—  S  has  the  slope  2  ; 

(d)  the  line  Sx-4y-k=0  has  its  :c-interce"pt  =  3  ; 

"^  (e)  the  line  3x  —  4^•?/+5  =  0  has  its  ^/-intercept  =  —  3 ; 

(/)  the  line  4:X—  Sy  +  k  =  0  is  distant  5  units  from  the  origin. 
A?is.     (a)  k=-2;     (c)  k  =  S  ;     (e)  k=-j%. 

V  4.    Write  the  equation  of  the  system  of  lines  perpendicular  to 
ij  y  =  mx  +  b. 

5.  Find  the  equation  of  the  system  of  lines  whose  normal  axis  has 
the  slope  —  f"^. 

6.  Find  the  equation  of  the  system  of  lines  whose  normal  axis  has 

the  inclination  — - . 
6 


THE   STRAIGHT   LINE  59 

7.  "What  is  the  equation  of  the  system  of  lines  tangent  to  a  circle 
having  the  center  at  the  origin  and  the  radius  10  ? 

41.   Systems  of  Lines  through  the  Intersection  of  Two  Given 

Lines.  —  Suppose  it  is  desired  to  find  the  equation  of  the 
system  of  lines  through  the  intersection  of  two  given  lines. 
One  way  of  doing  this  would  be  to  solve  the  two  given 
equations  for  the  point  of  intersection  and  to  use  the  point 
slope  equation.  A  method  more  expeditious  and  of  great 
theoretical  importance  follows. 

Theorem.  Let  /j  =  0  and  1^  =  0  be  the  equations*  of  two 
intersecting  lines  I^  and  L,  and  let  k  be  an  arbitrary  constant. 
TJien  J^-{-  k  '  L  =  0  defines  a  system  of  lines  passing  through 
the  intersection  of  the  given  lines. 

Proof.  We  first  observe  that  l^-]-  k  •  L  =  0  defines  a 
system  of  straight  lines  with  k  as  a  parameter,  since  it  is 
obtained  by  adding  two  first  degree  equations. 

Let  the  point  of  intersection  P  have  coordinates  (a,  b). 
By  the  locus  definition  the  equations  1^  =  0  and  1^  =  0  are 
both  satisfied  by  x  =  a  and  y  =  b,  since  P{a,  b)  lies  on  both 
curves.  That  is,  the  quantities  li(a,  b)  and  ^(a,  b)  are  both 
zero.     Therefore  ?i(o,  b)-\-  k  •  /^(ft,  b)=  0  for  all  values  of  k. 

Hence  every  line  of  the  system  Z^  -f  ^'  •  ?,  =  0  passes 
through  P,  since  the  coordinates  of  P  satisfy  the  equation.f 

*  The  equation  h  =  0  stands  for  an  expression  of  the  form 
Ax-{-B7j  +  C  =  0, 
and  would  more  properly  be  written  h{x,  y)=0,  but   the  variables  are 
omitted  for  convenience. 

t  We  can  also  prove  that  every  line  passing  through  P  belongs  to  the 
system.  Let  /g  =  0  be  another  line  through  P,  and  let  Pi(xi,  y{)  be  some 
other  point  on  l^.  If  the  coordinates  of  P^  be  substituted  in  li  +  k  -12  =  0, 
we  obtain  an  equation 

h{^\,  2/i)+A--/2(a^i,  2/i)=0 
in  which  k,  being  the  only  unknown, «.  an  be  found.    For  this  value  of  k,  the 
locus  of  the  equation  l^  +  k  •  1-2  =  0  passes  through  Pj,  since  its  coordinates 
make  the  left  hand  member  equal  to  zero.     As  it  already  passes  through  P, 
it  must  be  identical  with  l^,  having  two  points  in  common  with  it. 


60  ANALYTIC   GEOMETRY 

Example.  —  Find  the  line  containing  the  point  (2,  3)  and  the  inter- 
section of?i:  Sx  +  y  —  5  =  0  and  Zo:  x  +  2  y  —  S  —  0. 

The  system  Zi  +  A*  ■  Zo  =  0  contains  the  intersection,  and  k  may  be 
found  for  tlie  assumed  point  by  substituting  the  coordinates  of  that 
point  in  the  equation  of  the  system,  thus  : 

Sx  +  y -6  +  k{x  +  2y-3)=0 
becomes  6  +  3  -  5  -f  A:(2  +  6  -  3)  =  0, 

and  A:  =  —  4  ;  from  which  we  have  the  result  11  x  —  3  ?/  —  13  =  0. 

Verification.  Solving  the  given  equations  simultaneously,  the  inter- 
section is  (I,  f).  These  coordinates  and  the  coordinates  (2,  3)  both 
satisfy  the  new  equation. 

The  above  proof  will  apply  to  any  two  equations  of  the 
form  f(x,  y)  =  0,  as  well  as  to  two  linear  equations.  The 
general  principle  may  be  stated  as  follows : 

Tlie  equation  f(x,  y)-\-k  -  g(x,  y)  =  0  defines  a  system  of 
curves  passing  through  the  intersections  of  the  curves  defined  by 
the  two  equations  fix,  y)  =  0  and  g(x,  y)  =  0. 

PROBLEMS 

1.  Find  by  the  method  of  the  example  given  above  the  equation  of 
the  system  of  lines  passing  through  the  intersection  of  Sx  —  2y  =  16 
and  S  x  +  6  y  =  I.  Find  the  particular  line  of  the  system  passing 
through  (—  5,  4). 

i-  2.    Find  the  line  of  the  above  system  which  has  the  slope  |. 

Hint.  —  The  equation  in  k  is  3  a;  —  2  t/  —  16  +  A--(3  a;  +  6  ?/  —  1)  =  0, 
or  (3^•  +  3)x  +  (6A;-2)^/-fc-16  =  0. 

Solve  this  equation  for  y  and  the  coefficient  of  x  is  the  slope. 

V   3.    Find  the  line  of  the  above  system  which  is  parallel  to  the  a;-axis ; 
also  the  line  which  is  parallel  to  the  ?/-axis. 

.  4.    Find  by  the  method  of  the  illustrative  example  the  equations  of 
the  lines  determined  by 

V  (a)  P(4,  3)  and  the  intersection  ol  'ix—  2  y  =  7  and  x  —  2y  =  S  ; 
(6)   P(3,  0)  and  the  intersection  ofSx—2y  =  —  12  and  Sx  +  2y=0; 

(c)  P(—  5,  0)  and  the  intersection  oi  Sx  +  2y  =  o  and  x  =  6  ; 

(d)  P(—  4,  1)  and  the  intersection  of  x  —  y-\-6  =  0  and  x+y— 5=0. 

5.  Find  by  the  method  of  Ex.  2  two  lines  of  the  system  containing 
the  intersection  of  the  first  pair  of  lines,  one  parallel,  the  other  perpen- 
dicular to  the  third  line  : 

(a)  3  x  —  5  y  +  5  =  0,  5  X  +  4  2/  =  10,  and  x  +  y  =  6; 


THE   STRAIGHT   LINE  61 

(b)  X  +  y  =  0,  X  -  y  =  0,  and  2  X  +  S  y  +  6  =  0  ; 

(c)  6x  —  5y  +  i  =  0,x  +  y  =  l,  and  x  =  S; 

(d)  6  X  -  3  y  +  9  =  0,  4:  X  +  3  y  =  10  and  bx-9y  +  10  =  0. 

6.  Solve  Problem  4  (a)  by  finding  the  point  of  intersection  and 
using  (7). 

7.  Find  and  plot  the  equations  of  two  curves  which  pass  through  the 
intersections  of 

(a)  ?/2  =  8r/:+16,  (6)  x^  +  y^- =  13, 

2/2  =  4  X  +  16  ;  x2  -  2/2  -  5. 

42.  Plotting  by  Factoring.  —  Theorem.  If  1^  =  0  and  Zg  =  0 
define  the  lines  Ii  and  l^,  the  locus  o/  ?i  •  /^  =  0  consists  of  these 
lines. 

Proof.  This  follows  directly  from  the  locus  definition. 
For  the  coordinates  of  a  point  (a,  b)  on  either  line,  say  l^, 
satisfy  its  equation  l^  =  0.  But  if  li(x,  y)  vanishes  for  x  =  a 
and  y  =b,  the  product  of  li(x,  y)  and  /oC^j  y)  must  vanish  for 
the  same  values,  i.e.  (a,  h)  lies  on  the  locus  of  Z^  •  /,  =  0. 

Conversely,  if  (a,  h)  is  on  this  locus,  one  of  the  factors 
l^(x,  2/),  ?2(^j  y)  must  vanish  for  x  =  a,  y—h.  Hence  (a,  h) 
lies  on  the  line  defined  by  the  equation  formed  by  setting 
this  factor  equal  to  zero. 

This  proof  is  made  clearer  by  application  to  a  particular 
example. 

Consider  the  equation 

x^  —  ^  y"^  —  X  —  2  y  =  0. 
Factoring,  we  have 

{x^2y)(x-2y-l)  =  0. 

Now  the  coordinates  of  any  point  on  the  line  x-\-2  y  =  0,  as 
(2,  —  1)  make  the  first  factor  zero  and  hence  the  product  is 
zero.  Therefore  all  points  on  .^'  +  2  ?/  =  0  lie  on  the  locus  of 
the  given  equation.  The  same  is  true  of  x  —  2  y  —  1  —  0. 
Conversely,  if  a  point  lies  on  the  locus  oi  x^  —  1  y-  —  x  —  2  y =0, 
its  coordinates  satisfy  the  equation  and  must  make  at  least 
one  of  the  factors  of  the  left-hand  member  zero.     Therefore 


62  ANALYTIC   GEOMETRY 

all  points  of  the  given  locus  belong  to  one  of  the  lines 
x-\-2y  =  0,x-2y-l  =  0. 

This  reasoning  a23plies  to  all  equations  of  the  form 
f(x,  y)  =  0,  whether  they  deline  straight  lines  or  not,  although 
it  is  useful  chiefly  for  linear  factors.  It  leads  to  the  follow- 
ing rule. 

Rule  for  Plotting  by  Factoring.  —  Transpose  all  terms  to  the 
first  member.  Factor  as  far  as  possible  ;  set  each  factor  equal 
to  zero,  ayid  plot  the  resulting  equations  on  the  same  axes. 

PROBLEMS 

1.  Plot  by  factoring  the  pairs  of  lines  defined  by  the  following 
equations : 

(a)  a:2_  9?/2  =  0; 

tyj t</.      (c)  x^  -  8  x^y  -4:xy^  =0; 
y^       Id)  x'^—  x  —  y^  -\-y  =  0  ; 

(e)  4x-  +  4xy  +  y^  +  2  x  +  ?/  -  2  =  0  ; 
(/)  x^  +  2x+l-y'^  +  Qy-9=0; 
(9)  y^  +  Sy  +  2  =  0; 
(h)  x--iy^  -  x-2y  =  0. 

K^  2.    Draw  the  locus  defined  by  : 

^  .  /  /  .  (a)  5  xy^  =  7  x^y  ;      (b)  8  x^y  —  5  xy^  =  0. 

-"■jr  Find  the  area  of  the  triangle  defined  by : 
(a)  x^y-2xy^  +  Sxy  =  0; 
(6)   (x  +  S)(xy-y^)=0; 
(c)  x^  —  6x^-  xy^  +  6  ?/2  =  0. 

4.  The  sides  of  a  quadrilateral  are  defined  by  y'^  —  A  xy -\- ^  x'^ — 
^y  -\-  Qx  =  0  and  x^  +  4  x  +  3  =  0.     I'rove  that  it  is  a  parallelogram. 

5.  Plot  the  locus  of  the  equation  : 

(a)   (x2  +  ?/2  -  25) (x2  -  2/2  _  25)  =  0  ;     (6)  o-^y  +  xy^  =  16  xy. 

6.  Show  that  AxJ^  -\-  Bx  -\-  C  =  0  defines  a  pair  of  lines  parallel  to 
the  y-axis  if  B^  —  "iAOO,  a  single  fine  if  B^  —  i  AC  =  0,  and  an 
imaginary  locus  if  ^  —  4^C<0. 

7.  Show  that  the  locus  of  Ax-  +  Bxy  +  Cy^  =  0  is 

(a)  a  pair  of  intersecting  lines  if  B^  —  iAOO  ; 
(6)  a  single  line  \i  B'^  ~  4  AC  =  0  ; 
(c)  imaginary  ii  B-  -  4  AC <0. 


THE   STRAIGHT   LINE  63 

8.  Find  the  equations  of  the  bisectors  of  the  angles  between  the 
lines  defined  hy  oc^  —  4y'^  i-  S  x  +  6  y  =  0. 

9.  The  sides  of  a  parallelogram  are  y^  -  6  y  =  0  and  (y  —  2x)'-{- 
6(y  —  2x)  =  0.  Show  that  the  perpendiculars  upon  a  diagonal  from 
opposite  vertices  are  of  equal  length. 

10.  Generahze  and  solve  Problem  9. 

11.  The  vertices  of  a  triangle  are  (0,  0),  (5,  0),  (0,  ~  3).     Write 
an  equation  defining  its  three  sides.     Ans.  3  x^y  —  bxy-  —  \bxy  =  0. 


X 


X-  1 


CHAPTER   III 


THE    CIRCLE 

43.   The  Circle  Equation.  —  While  the  straight  line  equa- 
tion is  expressed  in  several  standard  forms,  depending  on 

the  choice  of  the  geometric 
constants  determining  it,  we 
use  only  one  such  form  of 
the  circle  equation  in  rec- 
tangular coordinates.  The 
constants  used  in  determin- 
ing the  circle  are  the  coor- 
dinates of  its  center  and  the 
length  of  its  radius. 

Let  the  coordinates  of  the 
center  of  the  circle  be  (li,  Ji),  the  length  of  the  radius  r,  and 
the  coordinates  of  the  point  tracing  the  circle  (;x,  y).  By 
the  distance  formula, 


r 

1 

^(^,y) 

0 

jc 

This  gives 


(13) 


r=V(x-h)--\-(y-k)\ 

-    (X  -  hf  +  (y  -  ky  =  r' 
as  the  standard  form  of  the  equation  of  the  circle. 

When  the  center  is  at  the  origin,  this  reduces  to  the  form 

x2  +  y'  =  f:  (13  a) 

PROBLEMS 

1.    Write  the  forms  which  the  equation  of  the  circle  takes  in  the 
following  cases  : 

(a)  center  on  the  x-axis  ;  (c)  circle  touching  the  a: -axis  ; 

(&)   center  on  the  y-axis;  (cZ)  circle  touching  the  ?/-axis  ; 

(e)  circle  touching  both  axes. 
64 


THE   CIRCLE 


65 


2.  In  each  of  the  following  cases  write  the  equation  of  the  circle  in 
the  standard  form  and  simplify  the  result ;  also  check  the  work  by 
finding  the  intercepts  and  drawing  the  circle  : 

v/  (a)  center  at  (3,  5)  and  radius  5  ; 

>C(6)  center  at  (—  8,  —  6)  and  radius  10  ; 

(c)  center  at  (4,  —  3)  and  radius  5  ; 

(d)  center  at  (—  8,  6)  and  radius  8; 

(e)  center  at  (6,  0)  3.nd  radius  6. 
Ans.     (a)  a;2  +  y2  _  6a-  -  10  ?/  +  9  =  0. 

yC  3.    Write  the  equations  of  the  circles  satisfying  the  following  condi- 
tions and  draw  the  figure  in  each  case : 

(a)  tangent  to  x  axis,  radius  3,  abscissa  of  center  6  ; 

(b)  tangent  to  y  axis,  radius  4,  ordinate  of  center  —  5; 

(c)  tangent  to  both  axes,  radius  16  ; 

(d)  passing  through  origin,  radius  5,  abscissa  of  center  —4  ; 

(e)  passing  through  origin,  radius  13,  abscissa  of  center  —  5; 
Rw,-i)0  passing  through(2,  2),  radius  5,  ordinate  of  center  —  1  ; 

<^ '    '^  Ans.     (a)  x2  +  2/2  _  12  X  ±  6  2/  +  36  =  0  ; 

(d)  x'-  +  i/+    8x±6y=0. 

4,    Prove  that  an  angle  inscribed  in  a  semicircle  is  a  right  angle.     . 

Solution.  —  Let  the  radius  of  the  circle  be  r;  take  the  diameter  as  the 
X-axis  and  the  center  as  the  origin.  Let  P  {x,  y)  be  any  point  on  the  cir- 
cumference.    We  have  to  prove  that  BPC  is  a  right  angle. 

Now  the  coordinates  of  B  are  (—  r,  0) 
and  of  C,  (r,  0).  Hence  by  the  slope  for- 
mula, the  slope  of  BP  is  -^  ;  that  of  CP 


X—  r 
have 


mm  =    ,  -^ 

r.2 


x-\-  r 
Calling  these  slopes  m  and  m',  we 

/2_ 

a;^  —  r'^ 

But  P(x,  y)  is  on  the  circle.     Therefore, 

x2  _|_  y-i  —  p2^   or  ?/2  =  /-"^  —  :c2.     This  give  s 

,      r2  —  x2  -, 

mm'  =  — ■  =  —  1, 

3.2  —  f>% 

which  proves  that  BP  and  CP  are  perpendicular,  or  that  BPC  is  a  right 
angle. 

v^  5.   Prove  without  the  use  of  slopes  that  an  angle  inscribed  in  a 
emicircle  is  a  right  angle. 


6.  Prove  that  a  perpendicular  from  any  point  on  the  circumference 
of  a  circle  to  a  fixed  diameter  is  a  mean  proportional  between  the  seg- 
ments into  which  it  divides  the  diameter. 


66  ,     }  ANALYTIC   GEOMETRY 

,■(/>' 

7.  Prove  that  angles  inscribed  in  the  same  segment  of  a  circle  are 
equal. 

Hint.  —  Take  the  chord  whose  ends  are  (c,  d)  and  (—  c,  d)  and  use  the 
angle  formula. 

8.  Prove  that  a  line  from  the  center  of  a  circle  bisecting  a  chord  is 
perpendicular  to  it. 

Hint.  —  Let  the  ends  of  the  chord  be  (—  r,  0)  and  (b,  c). 

44.  The  General  Equation  of  the  Circle.  —  If  we  expand  the 
standard  form  (13),  we  get  ^\^     Y~Kf/"^ 

This  is  of  the  form  i^ 

^^...^A..--^    x'^  +  y'  +  Dx+Ey^F  =  0,  (14) 

which  may  be  called  the  general  form  of  the  circle  equation. 
Conversely,  this  form  represents  a  circle,  since  by  complet- 
ing squares  it  is  possible  to  reduce  it  back  to  form  (13), 
which  we  know  is  the  equation  of  a  circle  of  center  (Ji,  h) 
and  radius  r.     (Xote  exception,  §  45.) 

The  general  equation  of  the  second  degree  has  the  form 

Ax"  +  Bxy  +  Cy''  +  Dx  +  Ey  +  F=0. 

Comparison  with  (14)  shows  that  this  can  be  reduced  to 
form  (14)  if  B  —  0  and  A  =  C,  by  dividing  by  A.  Hence 
we  have  the  theorem  : 

An  equation  of  the  second  degree  in  two  variables  defines  a 
circle  when,  and  only  when,  the  coefficients  of  x^  and  y"^  are 
equal  and  the  term  in  xy  is  ynissing. 

45.  Identification  of  Center  and  Radius.  —  To  find  the  cen- 
ter and  radius  of  a  circle  when  the  equation  is  given,  reduce 
the  equation  to  form  (14). 

Example.  — Find  the  center  and  radius  of  the  circle 

2  ic2  -  8  x  +  2  ^2  4-  6  ?/  -  21  =  0. 
Dividing  by  2  and  completing  the  squares,  we  have 

a;-2  _  4  x  -h  4  +  2/2  +  3  ?/  +  I  =  -y-  +  4  -h  f , 
or  {x-2y~+{y  +  iy=3-^-. 


THE   CIRCLE  67 

This  is  of  form  (13)  and  therefore  represents  a  circle  whose  center 
is  (2,   —  I)  and  whose  radius  is  ^V67  =  4.1  approximately. 

Degenerate  Forms.  —  Sometimes  on  completing  the 
squares  the  term  corresponding  to  r"^  is  negative  or  zero. 
In  the  former  case,  it  may  be  shown  by  the  methods  of  §  26 
that  the  locus  is  imaginary  and  we  say  that  the  graph  is  an 
imaginary  circle.  In  the  latter  case,  the  radius  of  the  circle 
is  zero,  and  the  graph  is  called  a  point  circle.  Such  forms 
are  sometimes  known  as  degenerate  forms. 

46.  Special  Forms.  —  Comparing  the  general  form  (14) 
and  the  expansion  of  the  standard  form  (13)  in  §  44,  we  see 
that        D  =  -2  7i,E  =  -2k,  and  F=h^+  A:^  _  ^,2, 

The  first  two  of  these  relations  enable  us  to  determine  the 
center  by  inspection.  They  also  enable  us  to  determine 
the  following  special  forms  : 

ii  D  =  0,  h  =  0  and  the  center  is  on  the  ?/-axis  ; 

ii  E  =  0,  k  =  0  and  the  center  is  on  the  a;-axis ; 

if  Z)  =  0  and  E  =  0,  the  center  is  at  the  origin ; 

if  i^"  =  0,  /i^  +  k^  =  r-,  and  the  origin  is  on  the  circumfer- 
ence. 

Exercise  1 .  Prove  by  considerations  of  symmetry  that  if  Z)  =  0, 
the  center  is  on  the  ?/-axis. 

PROBLEMS 

iX^l.  Prove  that  each  equation  defines  a  circle.  Also  determine  the 
center  and  radius  and  draw  each  circle. 

f,\      (a)  (a;  +  3)2+(y_4)2  =  25;    v'i:^" 
\^)  Xb)  (a;  +  4)2+(y-5)2  =  0; 

'/^y-^    j^yt^)   15  x^  +  lb  y'^  =  o  X  ; 
^  y  Sy</)  x2  +  ?/2  +  2x+8y  +  l=0; 

j;^  ■  (^)  2  x2  +  2  y2  _  16  X  -  20  ?/ =  0  ; 

\^Qi)  x2  +  2/2  +  5cc-3?/-i  =  0; 

V(*)  x2  +  ?/2_24a;  +  102/  +  169  =  0; 

^(j)  a:2-f  y2  +  2x  +  2?/  +  2  =  0. 


68  ANALYTIC    GEOMETRY 

2.  Determine  by  inspection  which  of  the  circles  in  Problem  1 

(a)  have  their  centers  on  the  .r-axis  ; 
(6)  have  their  centers  on  the  ?/-axis  ; 
i      .y  (c)  pass  through  the  origin. 

3.  Determine  by  inspection  the  centers  of  the  circles  in  Problem  1. 

4.  "Write  the  equations  of  the  circles  symmetrical  to  those  of 
Problem  1  with  respect  to  the  :c-axis  and  in  each  case  draw  both  circles. 

5.  In  the  system  of  circles  defined  by  (x  — 4)"^  +  («/  — 3)-  =  ^•, 
discuss  the  center,  radius,  and  intercepts  for 

^•  =  25,  16,  9,  4,  1,  0,  -1. 

6.  For  what  values  of  ^•  do  the  following  equations  have  a  locus  ? 
(a)  'x2  +  2/2  +  ^•^/  =  -  8  ;  (6)  x^  +  y^  ^.  )^y  =  -  6._ 

Am.   (a)  ^•  >  4  \/2. 

7.  Find  the  values  of  k  for  which  each  of  the  equations  of  Problem  6 
defines  a  point. 

/8.    Find  the  points  of  intersection  and  the  equation  of  the  common 
chord  of  the  circles : 

V{a)   cc2  +  y-2  +  8  2/  =  20,  (6)    x^  +  ?/2  +  4  a;  -  60  =  0, 

:c2  +  j/2  _  8  .c  =  4  ;  x-2  +  ?/2  _  4  ?/  -  36  =  0. 

9.  (a)  The  point  (|,  \)  bisects  a  chord  of  the  circle  x-  +  y-  =  25. 
Find  the  equation  of  the  chord  and  its  length. 

(6)  The  same  for  (4,  7)  and  a:2  +  ?/2  _  4  :>:  -  6  ?/  -  12  =0. 

Ans.     (a)  7  s:  +  ?/  -  25  =  0,  5\/2  ; 
(6)  a;  +  2?/-18  =0,  2V5. 

10.  Find  the  equation  of  a  circle  tangent  to  the  line  4x  +  3?/  —  7  =0 
and  having  its  center  at  the  origin. 

11.  Find  the  locus  of  the  vertex  of  a  right  triangle  which  has  the 
ends  of  its  hypothenuse  at 

(a)     (2,  3)  and  (-  2,  6) ;  (6)     (-  4,  3)  and  (0,  5). 

General  Hint.  —  In  each  of  the  following  locus  problems,  the  directions 

of  §  16  as  to  choice  of  axes  should  be  followed.  After  deriving  the  equation, 
the  locus  should  be  identified  and  drawn,  and  its  relation  to  tlie  given 
data  stated. 

12.  {a)  The  distance  between  two  fixed  points  is  2  c.  Find  the 
locus  of  a  point  moving  so  that  the  sum  of  the  squares  of  its  distances 
from  the  points  is  4  d^. 

(6)  Solve  the  same  problem  when  the  sum  of  the  squares  is  any 
constant,  as  k.     Ans.     The  locus  is  a  circle,  whose  center  is  midway 


THE   CIRCLE  69 


between  the  given  points  and  whose  radias  is  -^ — -  •     If  ^•  <  2  c^, 

there  is  no  locus. 

^3.   The  base  of  a  triangle  is  fixed  in  length  and  position.     Find  the 
locus  of  the  opposite  vertex  if 

—7    1 

(a)  the  vertical  angle  is  90°  ; 
y^6)  the  vertical  angle  is  45° ; 

(c)  the  vertical  angle  is  any  constant ; 

(d)  the  median  to  one  of  the  variable  sides  is  constant. 

14.  The  ends  of  a  straight  line  of  constant  length  rest  on  two  per- 
pendicular lines.     Find  the  locus  of  the  middle  point. 

15.  Find  the  locus  of  a  point  whose  distance  from  one  fixed  point 
is  k  times  its  distance  from  another. 

16.  Find  the  locus  of  a  point  if  the  sum  of  the  squares  of  its  dis- 
tances from  (a)  the  sides,  (6)  the  vertices,  of  a  given  square  is  constant. 

17.  From  one  end  of  a  diameter  of  a  circle  whose  radius  'S  r  chords 
are  drawn  and  produced  their  own  length.  Find  the  locus  oi  the  ends 
of  these  lines. 

/is.    From    the    point    (- r,  0)    chords    are    drawn    in   the    circle 
a;2  +  y-2  =  r^.     Find  the  locus  of  their  mid-points. 

47.  The  Equation  of  the  Circle  Derived  from  Three  Condi- 
tions. —  The  correspondence  between  the  geometrical  condi- 
tions necessary  to  determine  a  circle  and  the  algebraic 
conditions  necessary  to  determine  its  equation  are  analogous 
to  those  discussed  for  the  straight  line.  The  number  of  in- 
dependent constants  in  the  general  linear  equation' 

is  two,  for  if  we  divide  each  term  by  A,  we  have  the  form 

x+B'y-hC'  =  0. 

Geometrically,  a  straight  line  is  determined  by  two  points, 
or  in  general  by  two  geometric  conditions.  Both  forms  (13) 
and  (14)  of  the  circle  •  equation  involve  three  arbitrary  con- 
stants. Geometrically  a  circle  is  determined  by  three  points 
not  on  a  straight  line.     A  circle  may  be  determined  in  other 


70 


ANALYTIC    GEOMETRY 


ways  than  by  the  condition  of  passing  through  three  points, 
but  the  determining  condition  is  in  any  case  threefold. 

To  derive  the  equation  when  the  geometric  conditions  are 
given,  it  is  necessary  to  express  these  conditions  in  three 
equations  involving  h,  k,  and  ?-  (or  D,  E,  and  F)  and  solve 
them  simultaneously.  The  method  is  illustrated  by  the 
following : 

Example.  —  Find  the  equation  of  the  circle  determined  by  (1,  7), 
(8,  6),  and  (7,  -1).    ^  ,    I    ;.       ,    , 

Each  pair  of  these  coordinates  tnust  satisfy  the  circle  equation  (13). 
Hence  (1  -  h)'^  +  (7  -  k^  =  r^  \   i*>^-^^  v  ^. 

(7  _  A)2  +  (_  1  _  A,-)^  ^  ^2,^  V^' 

Solving  these  equations  simultaneously,  /i  =  4,  A:  =  3,  and  r  =  5. 
Therefore  the  equation  is  {x  —  4)^  +(?/  -  3)-  =  5'^,  which  reduces  at 
once  to  x^  +  ?/2  —  8  x  — ^y  =  0. 

We  may  also  solve  the  problem  by  using  form  (14),  which  gives 

l  +  49  +  Z)  +  7£'  +  i^=0, 

64  +  36  +  8Z)  +  6E'+i^  =  0, 

Solving,  i>  =  -  8,  E=-6,  and  F  =  0. 

A  convenient  graphical  check  on  either  solution  is  to  draw 
the  circle  with  compasses  and  observe  whether  or  not  it 
passes  through  the  three  given  points. 


48.   Length  of  a  Tangent. 

—  Let  t  be"  the  length  of  the 

tangent  P^T,  let  (h,  k)  be  the 

coordinates  of  the  center  O, 

rj^x'y^)    and  let  r  be  the  radius.    Then 

Using  the  distance  formula, 
this  becomes 


1^  =  (^x,-hy  +  {y,-ky-r\ 


(15) 


THE   CIRCLE  71 

It  is  observed  that  the  expression  for  t  is  the  same  in 
form  as  the  equation  of  the  circle,  with  the  coordinates  of 
Pi  substituted  for  the  variables. 

Hence,  to  find  the  length  of  a  tangent  from  a  given  point  to 
a  given  circle,  transpose  all  terms  of  the  given  equation  to  the 
first  member,  substitute  in  the  expression  thus  formed  the  coor- 
dinates of  the  given  point,  and  extract  the  square  root. 

ISTote  the  similarity  of  this  to  the  formula  for  the  distance 
from  a  line  to  a  point. 

Since  the  right-hand  member  of  (15)  is  the  same  in  form 

as  the  standard  equation  of  the  circle,  it  may  also  be  written 

f  =  X,'  +  j/r  -f  Dx,  +  Ey,  -h  F.  (loo) 

Example.  —  Find  the  length  of  the  tangent  from  (5,  6)  to  the  circle 

x"  +y-  —  ix  -\-  6y  —  S  =0. 

Using  formula  (loa),  we  have 

f2  =  52  _[.  6-2  -  4  •  5  +  6  .  6  -  3  =  74. 

t  =  VtT 


PROBLEMS 

/T    Find  the  equation  of  the  circle  determined  by  the  points : 
<^)    (-1,2),  (4,  2),  (-3,4);  (d)    (- 5,  0),  (0,  4),  (2,  4)  ; 

(b)  (0,  0),  (6,  0),  (0,  -  8)  ;  (e)   (2,  0),  (0,  0)  ,(-  2,  -  2)  ; 

(c)  (1,  1),  (1,  3),  (9,  2)  ;  (/)    (6,  0),  (-  2,  4),  (0,  -  6). 

^   Find  the  equation  of  the  circle  which 

(a)  has  the  center  (3,  4)  and  passes  through  (4,  —  3)  5 

(b)  has  the  center  (r,  0)  and  is  tangent  to  the  ^/-axis  ;  ^      --^^. 

(c)  passes  through  (2,  0)  and  (6,  0)  and  is  tangent  to  the  ?/-axis ; 
^d)  has  the  line  joining  (6,  2)  and  (—  8,  6)  as  a  diameter; 

(p)  passes  through  (3,  2)  and  ( —  2,  —  1)  with  center  on  the  x-axis  ; 
(/)  circumscribes   the  triangle   whose  sides  are  the  lines  x  =  5, 
y  =-S,  x-2y=l  ; 

(g)  passes  through  (9,  8)  and  is  tangent  to  both  axes. 

Ans.    (6)  y'^  =2rx  —  x^  ; 

(e)  5x2+  5?/2-8x  =  41; 

ig)  x^  +  1/2  _  lOx  -  10  ?/  +  25  =  0  or  x2  +  ?/2  -  58  X  -  58  ?/  + 
841  =  0. 


72 


ANALYTIC   GEOMETRY 


♦^^3.   Find  the  equation  of  the  circle  which 
1^-/  "  (^)  is  tangent  to  the  line  x  -\-  2y  —  10  =  0  and  passes  through  the 
^^^^oints  (3,  3)  and  (0,  0)  ; 

(6)  has  the  center  (1,  5)  and  is  tangent  to  the  line  Sx  —  4y-\-7  =  0', 

(c)  is  tangent  to  the  lines  2  x  —  Sy  =  0  and  3  a:  +  2 z/  —  13  =  0  and 
passes  through  the  point  (—  4,  6)  ; 

(d)  is  tangent  to  the  line  3a:  —  4?/  =  2at  the  point  (2,  1)  and  passes 
through  the  origin.  Ans.     (a)   (x  -  1)2  +  (y  -  2)^  =  5, 

or  (X--2^^)2+  (y +1)2=8/^5. 

(c)  (X  +  2)2  +(y-  3)2  =  13, 
or  (x  +  22)2+  (?/-7)2  =  325; 
^^-^  (d)  2x^  +  2y^  +  7x-24y  =  0. 

4.  Find  the  length  of  the  tangent  from  the  point  (6,  8)  to  each 
circle  of  Problem  1,  page  67. 

*^5.   Find  the  least  distance  from  the  point  (6,  8)  to  each  circle  of 
Problem  1,  page  67.  Ans.    (a)    v  97  —  5. 

6.  Find  the  coordinates  of  the  points  of  contact  of  the  tangents 
from  the  given  point  to  the  given  circle  : 

(a)  x^-\-y^-6x-6y-l  =  0,  (-4,4), 
(6)  x2  +  2/2=17,  (5,3). 

Hint.  —  The  coustruction  of  the  tangents  by  the  usual  geometric  process 
will  suggest  a  method  of  solution. 

49.  Common  Chord  of  Two  Circles.  —  To  find  the  common 
chord  of  two  circles  we  may  solve  the  equations  simul- 
taneously, to  get  the  points  of  intersection,  and  then  use  the 

two-point  form  of  the  line 
equation.  A  better  method 
will  now  be  illustrated. 

Let  the  two  circles  have 
the  equations 

x'+f—Sx-4.y-U=0,  (a) 
and  x''-{-7f—2o  =  0.    (b) 

Subtracting, 

Sx-\-4.y-{-19  =  0.    (c) 

By  the  theorem  of  §  41  the 

locus  of  (c)  contains  the  points  common  to  (a)  and  (b).     Since 

it  is  of  the  first  degree,  it  defines  a  straight  line.     Hence 

this  line  is  the  common  chord. 


e..^i9^^  PC  A-- 


THE   CIRCLE  73 


The  equation  of  the  common  chord  of  two  circles  can 
always  be  obtained  by  subtracting  the  equation  of  one  of 
them  from  that  of  the  .other,  as  above.  For,  since  noth- 
ing above  the  first  degree  appears  in  the  equation  of  a 
circle  except  the  characteristic  terms  x^  +  y"^,  which  are 
always  present,  these  will  cancel  by  subtraction,  leaving  a 
linear  equation,  whose  locus  contains  the  points  of  inter- 
section of  the  two  circles. 

Exercise  2.  Derive  the  general  equation  of  the  common  chord  of 
two  circles.  Ans.    (D—  D'yx +(E  —  E')y  +  (F—  F')=  0. 

Exercise  3.  Prove  that  the  common  chord  of  two  circles  is  perpen- 
dicular to  their  line  of  centers. 

Hint.  —  The  line  of  centers  of  two  circles  is  determined  by  the  points 


(h,  k)  and  (h',  k').    Hence  its  slope  is 


k-k' 


Compare  this  with  the 


y 


h—h' 

slope  of  the  common  chord,  found  by  solving  tlie  answer  to  Exercise  2 
for  y,  and  expressing  the  result  in  terms  of  h,  k,  h',  and  k'. 
-■y 
50.   Radical      Axis.  —  Con- 
sider   the    following    two  ex- 
amples. 

Example  1.  — Find  the  equation 
of  the  common  chord  of  the  circles 

(x  +  3)2  +  2/2  =  25 
and  (x  —  5)2  +  y^  =  4, 

and  the  intersections  of  the  circles. 

Applying   §  49,  the   equation   is 
found    to    be    16  x  =  37  ;    but   the 
points  of    intersection    are   imagi- 
nary, as  is  seen  by  inspection  of  the  equations,  since  the  sum  of  the 
radii  is  7,  while  the  distance  between  the  centers  is  3  +  5  =  8. 

Example  2.  —  Find  the  equation 
of  the  common  chord  of  the  circles 

(X  +  3)2  +  2/2  =  25 
and         (x  -  5)2  +  y^  =  9. 

Here  the  distance  between  the 
centers  is  the  same  as  the  sum  of 
the  radii  and  the  circles  are  tan- 
gent to  each  other  at  the  point 
(2,  0).     Also  we  have  by  subtrac- 


74 


ANALYTIC    GEOMETRY 


tion  X  =  2.  whose  locus,  passing  through  the  point  (2,  0)  and  being 
perpendicular  to  the  line  of  centers,  is  the  common  tangent. 

Thus  vre  see  that  the  application  of  the  rule  for  finding  a 
common  chord  gives  a  common  chord  when  the  two  circles 
intersect,  a  common  tangent  when  they  are  tangent,  and 
linally,  when  tliev  do  not  meet,  a  line  which  has  no  apparent 
significance,     ^^'e  shall  hnd,  however,  that  the  line  in  all 

three  cases  has  a  common 
property ;  viz.  it  is  the  locus 
of  the  point  from  which  tan- 
gents to  the  two  circles  are 
equal. 

To  prove  this,  let  t  and  f  be 
equal  tangents  from  P  to  the 
circles  C  and  C  respectively,  of 
which  the  equations  are 

x'-  +  f-  +  D:c  +  Ey-^F=0 
X'-  -f  ?/-  +  D'x  +  E'y  ^F'  =  0. 
From  §  48,  (15a),      t-  =  x'-  +  ?/-  +  Dx  +  Ey  -^  F. 
Also  r-  =  x"-  +  1/  +  D'x  +  E'y  -f  F'. 

Hence  f  -  t'-  =  (D  -  D')x  +{E-  E')y  j^F-F'. 

By  hypothesis,  t  =  f,  or  t-  —  t'-  =  0. 

Therefore,  (D-  iy)x -\- (E  -  E')y  -\- F  -  F'  =  0  (16) 
is  the  equation  of  the  locus  of  P. 

But  this  is  the  equation  obtained  by  eliminating  the  terms 
in  a;2  and  y-.     Hence  we  have  the  theorem : 

TJie  locus  of  the  point  from  which  the  tangents  to  two  circles 
are  equal  is  the  locus  of  the  equation  obtained  by  eliminating 
the  terms  in  a-^  and  y-  between  the  equations  of  the  circles. 

This  locus  is  called  the  radical  axis.  When  the  circles 
intersect,  it  is  their  common  chord ;  and  when  they  touch, 
it  is  their  common  tangent. 


and 


^ 


^ 


THE   CIRCLE  75 

PROBLEMS 

lA.    Find  the  equations  of  the  radical  axes  and  the  points  of  inter- 
section of  the  following  pairs  of  circles  : 

(a)        x2  +  ?/2_^4x  +  6  2/=7,                 id)  y?^y^\^x^-Zy-^2=^, 

a;24.2/2_2a:-6?/  +  5  =  0;  a;2  +  ?/2  +  3x  +  5?/-28  =  0; 

(6)         a;2+2/2_5x-5?/=-ll,            (e)  a:2_^y2_2x+4?/-4=0, 

a:2  +  y2_9^0;  2a:2  +  2i/2_4a^  +  6y_7^0; 

^c)       a;2  +  ?/2  +  8a:  +  18?/  =  l,                 (/)  a:2+7/H7x-5  2/  =  8, 

3a;2+32/2  +  20a:+44yzzll;  x2  +  ?/2+8a:-4?/=3  5. 

2.  Find  a  point  from  which  equal  tangents  may  be  drawn  to  each 
of  thertollowing  circles  : 

(a)  a;2_^y2  + 3^^4.2?/  =  6,  i^)        x2+y2^4^_32/^5^0, 

a;2  +  2/^  _  3 X  -  2  ?/  =  -  2,  2a:2-f-2?/24.5a:_6y+l  =0, 

a:2  +  2/2_4a;_8y  =_12;  332+^2^5  a._4y^5^0. 

3.  Prove  that  the  radical  axes  of  any  three  circles  taken  in  pairs 
meet  in  a  point  or  are  parallel. 

^     •^.   In  each  pair  of  circles  in  Problem  1  find  the  line  of  centers,  and 
/V  prove  that  it  is  perpendicular  to  the  radical  axis. 

51.  Systems  of  Circles.  —  The  geometric  constants  of  the 
circle  designated  by  U^  k,  and  r  determine  the  center  and 
radius.  If  two  of  these  constants  are  restricted  by  assign- 
ing conditions  to  them,  numerical  or  otherwise,  the  equation 
•  will  represent  a  system  of  circles  defined  more  or  less  com- 
pletely according  to  the  nature  of  the  conditions. 

Thus,  if  7i  =  2  and  k  =  4,  we  have  a  system  of  concentric 
circles  of  varying  radius.  Again,  it  h  =  k  and  r  =  10,  we 
have  a  system  of  equal  circles  whose  centers  lie  on  the  line 
X  —  y  =  0. 

Systems  having  the  Same  Radical  Axis.  —  If  we  use  the  theo- 
rem of  §  41,  we  find  that  C+k-  C  =  0,  where  k  is  any  constant 
and  C  =  0  and  C"  =  0  are  the  equations  of  two  circles,  will  define 
a  system  of  circles  passing  through  the  points  of  intersection 
of  G  and  C.  But  the  radical  axis  of  any  two  circles  passes 
through  the  points  of  intersection,  real  or  imaginary ;  hence 
this  equation  defines  a  system  of  circles  having  the  same 
radical  axis. 


76 


ANALYTIC   GEOMETRY 


The  figure  below  illustrates  the  system  of  circles  having 
the  same  radical  axis  as       2  1   ,,2  1     4  o'  _  qa  —  0 


and 


0. 


x"  +  ?/2  -  21  X  +  54 

These  meet  at  the  points  (6,  6)  and  (6,  —  6) ;  the  equation 
of  the  radical  axis  is  x=  Q.     The  equation  of  the  system  is 

^2  +  ?/-  +  4  a;  -  96  +  kif  +  /-  -  21  a;  +  54)  =  0. 


/fc=-"xr 


:\i^ 


I  I  I  I 


4\     -is 


1    IN    I    I    II 


In'  the  figure  the  circles  drawn  in  heavy  lines  are  the  given 
circles  and  the  others  are  the  circles  corresponding  to 

A:  —  11   4    — J-    — 1    — 1    — A_3    _6 

Note  that  when  k  approaches   —  1,  the  circle  becomes  of 
infinite  size  and  approaches  as  a  limit  the  radical  axis. 


CHAPTER   IV 


THE   PARABOLA 


52.  Conies.  —  The  locus  of  a  point  ichich  moves  so  that  its 
distance  from  a  fixed  point  is  in  a  constant  ratio  to  its  distance 
from  a  fixed  line  is  called  a  conic. 

The  fixed  point  is  called  the  focus  and  the  fixed  line  the 
directrix.  The  line  through  the  focus  perpendicular  to  the 
directrix  is  called  the  principcd  axis.  The  constant  ratio  is 
called  the  eccentricity  and  is  represented  by  the  letter  e. 

There  are  three  kinds  of  conies,  classified  as  follows  : 

e  =  1,  the  parabola  ; 
e  <  1,  the  ellipse  ; 
e  >  1,  the  hyperbola. 

53.  The  Equation  of  the  Parabola.  —  Let  AB  be  the  direc- 
trix, F  the  focus,  and  DF  the  principal,  axis  of  the  parabola. 

By  the  definition  of  the 
parabola,  there  is  a  point  0, 
halfway  between  the  focus 
and  the  directrix,  which  lies 
upon  the  locus.  This  is 
called  the  vertex.  To  derive 
the  equation  take  the  prin- 
cipal axis  as  the  a:-axis  and 
the  vertex  as  the  origin. 
Call  the  distance  between 
the  directrix  and  the  focus 
p.  Then  F  has  the  coordi- 
nates [-^,  0  |,  and  the  directrix  has  the  equation  x  = 


D 

Y 

y"^ 

JS 

D 

0 

F(P/2,0)                          X 

1 

V 

1 
II 

\v 

A 

\^ 

2 


.\ 


c/ 


77 


78  ANALYTIC   GEOMETRY    . 

Let  P{x,  y)  be  any  point  on  the  locus.     Draw  NP  perpen- 
dicular to  AB,  and  join  FP.     By  definition  FP  =  XP, 


that  is,  yj(x-ijj+f  =  x+^. 

Squaring  and  reducing, 

y'  =  2px.  (17) 

This  is  the  standard  equation  of  the  parabola  when  the 

vertex  is  taken  as  the  origin  and  the  principal  axis  as  the 

a>-axis.     If  the  principal  axis  is  taken  as  the  ?/-axis  and  the 

vertex  as  the  origin,  the  variables  are  interchanged  and  the 

equation  is 

x'  =  2py.  (17a) 

If  the  focus  is  in  a  negative  direction  from  the  vertex,  the 
right-hand  members  of  these  equations  have  minus  signs. 

The  simplicity  of  these  equations  is  due  entirely  to  the 
choice  of  the  coordinate  axes.  If  the  axes  are  chosen  in 
any  other  way,  a  different  and  more  complicated  equation 
is  obtained,  but  the  curve  itself  is  unaltered.  Therefore  we 
shall  study -the  properties  of  the  parabola  and  other  conies 
by  means  of  the  simple  forms  and  reserve  the  general  equa- 
tions for  a  later  chapter. 

Exercise  1.     Derive  equation  (17a)  from  a  figure. 

54.  Discussion  of  the  Equation.  —  The  form  of  equation 
(17)  shows  that  the  parabola  is  symmetrical  with  respect  to 
the  principal  axis,  and  that  the  parabola  crosses  this  axis  at 
the  vertex  only. 

Since  negative  values  of  x  make  y  imaginary,  the  curve 
lies  wholly  to  the  right  of  the  vertex  and  has  no  point 
nearer  the  directrix.  It  recedes  indefinitely  from  both  axes, 
since  y  =  ±  V2  px  increases  indefinitely  with  x.  A  different 
parabola  is  obtained  for  each  value  assigned  to  2^- 


THE   PARABOLA 


79 


f 

Y 
L     J 

.^_- 

B 
X 

F 

[1 

L' 

V 

D        X 
A 

Exercise  2.  Derive  the  equation  of  the  parabola,  taking  the  focus 
at  the  left  of  the  directrix,  and  discuss  as  in  this  section. 

Ans.  y-  =—  2 px. 

55.  Latus   Rectum.  —  The 

chord  L'L  through  the  focus 
parallel  to  the  directrix  is 
called  the  latus  rectum.  By 
definition  of  the  parabola, 
the  distance  of  L  from  the 
focus  is  the  same  as  that 
from  the  directrix. 
Hence 

FL  =  LN=FI)  =  i) 

and  the  length  of  the  whole 
latus  rectum  is  2  p. 

56.  To  Draw  a  Parabola.  — Consider  the  equationa.'^  =  —  8y. 
This  is  of  the  form  x"^  =  —  2  py,  and  hence  is  the  equation  of 
a  parabola.  To  draw  the  curve  we  can  make  a  table  of  values 
as  usual,  or  proceed  according  to  one  of  the  two  following 
methods. 

Method  1.  From  the  form  of  the  equation,  2^j  =  8,  or 
J)  =  4.     Since  the  sign  before  2  p  is  minus,  the  curve  lies 

wholly  below  the  ic-axis. 

Hence  the  focus  is  f  0,  —  -^ 
V  2 

or  (0,  —  2),  and  the  equation  of 
the  directrix  is  2/  =  2.  Meas- 
uring off  p  =  4  to  the  right 
and  left  of  the  focus  gives 
the  ends  of  the  latus  rectum. 
These  together  with  the  vertex 
0  make  three  points  on  the  curve,  which  suffice  for  a 
sketch,  as  the  general  shape  of  a  parabola  is  known. 


80 


ANALYTIC    GEOMETRY 


Method  2.     If  a  more  accurate  graph  is  desired,  find  the 
focus   and  directrix  as  above.     Then  with   the   focus  as  a 

center    and    ^    v-arlina    -,.  \-? 


T 

A 

5 

-Y 

.    \ 

F 

"^^ 

r 

\ 

V 

a 


radius    r  >  ^- 

9 


draw  an  arc.  This  will  cut 
the  line  parallel  to  the  di- 
rectrix and  r  units  from  it  in 
two  points  which  lie  on  the 
parabola  by  definition.  By 
using  coordinate  paper  and 
varying  r,  as  many  points  as 


are  desired  mav  be  readilv  constructed. 


57.  Mechanical  Construction  of  the  Parabola.  —  Place  a  right 
triangle  with  one  leg  CE  on 
the  directrix.  Fasten  one 
end  of  a  string  whose  length 
is  CD  at  the  focus  F  and 
the  other  end  to  the  triangle 
at  D.  With  a  pencil  at  P 
keep  the  string  taut.  Then 
FP  =  CP ;  and  as  the  triangle 
is  moved  along  the  directrix 
the  point  P  will  describe  a 
parabola. 


PROBLEMS 

1.  "With  coordinate  paper  and  compasses  plot  a  parabola  having 

the  latus  rectum :  ,      , 

(a)  12;     {h)   16. 

2.  Find  the  coordinates  of  the  focus  and  the  equation  of  the  direc- 
trix of  each  of  the  following  parabolas  and  draw  the  figure  to  scale  : 

(a)2/2^8a;;  (c)  2  x^  +  5?/ =  0  ;  (e)?/2  =  6aa;; 


3.    What  is  the  latus  rectum  of  each  parabola  in  Problem  2  ? 


THE   PARABOLA  81 

4.    Write  the  equations  of  parabolas  satisfying  the  following  data : 
(a)  directrix  ?/  =—  8,  focus  (0,  8)  ; 
^/(b)  directrix  x  =  6,  focus  (  —  6,  0)  ; 
^(c)  directrix  y  =  8,  vertex  (0,  0)  ; 
{f^{d)  latus  rectum  12,  vertex  (0,  0)  ; 
^^ie)  vertex  (0,  0),  one  point  (3,  —  4). 
j^y^  Ans.    (a)  x^  =  32y. 

Derive  the  equation  of  the  parabola  when  the  principal  axis  is 
(a)  as  the  :/>axis,  with  the  origin  at  the  focus  ; 
(6)  as  the  y-axis,  with  the  origin  at  the  focus ; 
.l/^)  b^  the  a"-axis,  with  the  origin  on  the  directrix  ; 
A^)  as  the  ?/-axis,  with  the  origin  on  the  directrix. 

Ans.    (a)  2/2  =  2px  +p^; 
(c)   y^  =  2px  —  p^. 

6.  Find  the  locus  of  the  center  of  a  circle  passing  through  a  given 
point  and  tangent  to  a  given  line. 

7.  Find  the  equation  of  the  circle  circumscribing  the  portion  of 
the  parabola  x~  =  2py  cut  off  by  the  latus  rectum. 

Ans.    2x^  +  2y^  —  bpy  =  0. 

^8.  Find  the  points  of  intersection  of  the  parabolas  x^  =  2py  and 
y^  =  2px. 

9.    In  the  parabola  y^  =  2px  an  equilateral  triangle  is  inscribed 
with  one  vertex  at  the  origin.     Find  the  length  of  a  side. 

Ans.  4p  Vs. 

-XIO.    Show  that  the  distance  of  any  point  of  the  parabola  y"^  =  2px 
from  the  focus  is  a:  + -.     (This  is  called  the  focal  radius  of  the  point.) 

11.  Show  that  the  latus  rectum  of  any  parabola  is  a  third  propor- 
tional to  the  abscissa  and  ordinate  of  any  point  upon  it. 

12.  Find  the  area  of  the  right  triangle  inscribed  in  the  parabola 
y^  =  2px  which  has  a  double  ordinate  as  a  hypothenuse  and  a  vertex 
at  the  origin.  Ans.  ip'^. 

flZ.  One  end  of  a  chord  through  the  focus  of  a  parabola  which  has 
the  a;-axis  as  the  principal  axis  and  the  origin  as  the  vertex  is  the 
point  (—8,  8).     Find  the  coordinates  of  the  other  end. 


82 


ANALYTIC   GEOMETRY 


58.  The  Parabola  as  a  Conic  Section.  —  In  tlie  right  circular 
cone  0-AB  pass  a  plane  through  the  yertex  0  and  the 
diameter  of  the  base  AB.  It  will  be  perpendicular  to  the 
base  and  will  contain  the  elements  OA  and  OB,  and  also  the 

axis  and  the  center  of 
every  section  parallel  to 
the  base. 

Draw  a  plane  perpen- 
dicular to  the  plane  OAB 
intersecting  it  in  tlie  line 
MX  parallel  to  OA,  and 
cutting  the  surface  of  the 
cone  in  CXD.  Through 
P,  any  point  of  CXD, 
and  X  pass  planes  par- 
allel to  the  base,  inter- 
secting the  plane  OAB 
in  HX  and  FG,  which  are  the  diameters  of  the  circular 
sections  thus  formed. 

Now  QP,  the  intersection  of  the  }»lanes  FPG  and  CXD,  is 
perpendicular  to  OAB  and  hence  to  the  lines  FG  and  XM. 
Taking  X  as  the  origin  and  XM  as  the  ^axis,  the  coordinates 
of  P  are  x  =  XQ  and  y  =  QP. 


Then  i/=qP'=Fq 

By  elementary  geometry, 


QG. 


FQ  =  HX,  and  Sg  =  ^^,  or  QG  = 


NQ      OH' 


HX 
OH 


X. 


Substituting, 


if  = 


HX' 
OH 


•T. 


As  HX  and  OH  are  constants,  being  independent  of  the 
position  of  P,  ^^  may  be  taken  as  2  p.     Thus  the  section 

CXD  is  a  parabola. 


THE   PARABOLA 


83 


59.  The  Path  of  a  Projectile.  —  A  ball  is  thrown  horizon- 
tally from  a  point  above  the  ground.  Find  its  path,  neglect- 
ing air  resistance. 

Let  V  be  the  velocity  of  projection.  Take  the  point  of 
projection  as  the  origin  and  a  horizontal  line  through  it  as 
the  a^axis. 

Let  P{x,  y)  be  the  position 
of  the  ball  after  t  seconds.  If 
V  is  measured  in  feet  per  sec- 
ond, the  ball  will  have  moved 
horizontally  a  distance  vt  feet, 
that  is, 

x  =  OQ  =  vt. 

But  during  the  same  time,  by  a  principle  of  physics,  the  ball 
will  have  fallen  a  distance  i  gt"^  where  g  =  S2  nearly.     Hence 

Eliminating  t  between  these  equations. 


fi  — 


2^2 

-■ — y- 

9 


Therefore  the  locus  is  a  parabola  having  its  vertex  at  0  and 


its  latus  rectum  of  length 


Z  V' 


9 


The  figure  is  drawn  for 


-y  =  64  feet  per  second. 

Later  we  shall  find  that  the  path  of  a  projectile  hurled  in 
any  direction  is  a  parabola. 


PROBLEMS 

1.  The  altitude  of  a  right  circular  cone  is  10  inches  and  the  diameter 
of  the  base  is  8  inches.     If  the  vertex  of  a  parabola  constructed  as  in 

§  58  bisects  an  element  of  the  cone,  find  its  latus  rectum,   Ans.  —  \/29. 

29 

2.  If  two  parabolic  sections  are  made  as  in  §  58,  prove  that  their 
latus  rectums  are  proportional  to  the  distances  of  their  vertices  from 
the  vertex  of  the  cone. 


84  ANALYTIC    GEOMETRY 

3.  If  the  vertex  of  a  parabolic  section  is  made  to  approach  the 
vertex  of  its  cone,  what  is  the  limiting  form  of  the  parabola?  of  its 
equation  ? 

4.  A  bomb  is  dropped  from  an  airship  traveling  east  at  30  miles 
per  hour  at  a  height  of  1  mile.  Write  the  equation  of  its  path,  re- 
ferred to  the  starting  point  as  an  origin.  How  far  east  will  it  be  when 
it  strikes  the  ground  ?  Ans.   799  feet. 

5.  A  ball  was  thrown  horizontally  on  a  level  field  at  a  height  of 
5  feet  and  struck  the  ground  100  feet  from  the  starting  point.  What 
was  its  initial  velocity  ? 

"    Q.    An  aviator  flying  horizontally  at  45  miles  per  hour  wishes  to 
lit  a  target  on  the  ground.     He  estimates  his  height  above  the  ground 
at  1000  feet.     How  far  away  from  the  target  should  he  be  when  he 
drops  his  projectile  ?  Ans.    522  feet. 

7.  An  airship  which  flies  at  30  miles  per  hour  has  a  device  for  re- 
leasing a  projectile  when  it  is  directly  over  the  center  of  the  target. 
How  large  must  the  target  be  to  insure  hitting  it  when  the  height  -of 
the  airship  is 

(a)  ^  mile  ;  (6)  2  miles  ;  (c)   li  miles  '? 

Ans.    (a)    1130  feet  wide. 

'~  8.    Sketch  the  parabolas  satisfying  the  following  data  by  finding 
the  axis,  vertex,  and  ends  of  the  latus  rectum  : 

^        J     (a)  directrix  y  =  5,  focus  (4,  —  3)  ; 

X    /  Ans.    Axis  x  =  4,  vertex  (4,  1),  ends  of  latus  rectum 

/  (12,  -3)  and  (-4,  -3). 

y^^/^b)  directrix  x  =  -S,  focus  (—  1,  4)  ; 
^■j^^^^^ji^     (c)  directrix  y  =Q,  focus  (4,  —  2)  ; 
''  (d)  directrix  x  =  6,  focus  ( —  6,  —  6) . 

'^-'9.   Derive  the  equations  of  the  parabolas  of  Problem  8. 
,,       '    .  /  Ans.    (a)  x^—Sx+16y  =  0. 


10.    Show  that  the  circle  x-  —  2  ax  +  y'^  =  0  and  the  parabola  y'^  = 
2px  meet  in  but  one  point  unless  a>p. 

n.1.    Find  the  locus  of  the  mid-points  of  the  ordinates  of  the  parab- 
ola y^  =  2  px. 

12.    Find  the  locus  of  the  ends  of  the  latus  rectum  of  the  parabola 
y'^  =  2px  when  p  is  allowed  %o  vary. 


CHAPTER   V 
THE   ELLIPSE 


60.  Fundamental  Constants.  —  In  §  52  the  ellipse  was  de- 
fined as  a  conic  of  eccentricity  less  than  1. 

In  the  figure  let  AB  be  the  directrix,  F  the  focus,  and 
DF  the  principal  axis. 


. .p — 


r**-  — -« — >k a — Jy' 


Since  e  <  1,  there  are  two  ^ 

points  V  and  V  on  the 
principal  axis,  one  on  either 
side  of  the  focus,  satisfying  d 

the  definition  of  the  ellipse. 
These  are  called  vertices,  the 
point  C  midway  between  f^ 
them  is  called  the  center,  and  the  distances  between  Fand 
C,  and  F  and  C  designated  by  a  and  c  respectively.  Be- 
tween the  four  constants  a,  c,  p,  and  e  exist  certain  im- 
portant relations  which  we  now  proceed  to  derive. 

Applying  the  definition  of  the  ellipse  to  the  points  Fand 
V,  we  have 

FV'  =  eDV',  or  a-{-c  =  e(c-\-p  + a) 
and  VF=eDV,    ov  a  — c  =  e{c+p  — a).  ^ 

Adding  and  subtracting  these  equations, 

a  =  e(c-j-p),     or     c +  />  =  -,  (18) 

e 

and  c  =  ae.  (19) 

The  first  relation  gives  the  distance  between  the  center 
and  the  directrix  in  terms  of  a  and  e,  and  the  second  that 
between  the  center  and  the  focus. 

Exercise  1.  Show  how  to  find  Fand  F'  with  ruler  and  compasses, 
when  the  directrix,  focus,  and  eccentricity  are  given. 

85 


86 


ANALYTIC    GEOMETRY 


61.    The  Equation  of  the  Ellipse.  —  To  derive  tlie  equation 
of   the  ellipse  take  the  principal  axis  for   the  a:-axis  and 

the  center  for  the  origin. 
Then  F  has  the  coordinates 
(—  ae,  0)  and  the   directrix 

has  the  equation  it'  = 

e 

Let  P  be  any  point  of  the 

locus.      Then   FP=e'NP. 

By  the  distance  formula 


z> 

Y 

^^-r,?/) 

N 

(..^■' 

v\ 

iF{-ae,o)0 

Jd     jv'X 

> 

J 

II 

_»-..^^ 

b 

' 

d' 

FP  =  V(x  +  0£f  H-  2/^ 


and 

Substituting  these  values 


e 


V(a.'  +  ae)-  -{-  y"-  =  e[  x -\--\=  ex  -\- 

Squaring  and  collecting  terms, 

(l-e'-)x^  +  f-  =  a\l-e''). 


a. 


Dividing  by  a-(l 


e2),     ^  + y =1. 


Since  e  <  1,  the  quantity  a-{l  —  e-)  is  positive  and  <  «-. 
Calling  this  6-,  we  have  the  standard  form  of  the  equation  of 
the  ellipse  when  the  center  is  taken  as  the  origin  and  the 
principal  axis  as  the  x-axis,  namely  : 


1, 


(20) 


or 


62x'2  +  a-if  =  ci-h- 


62.    Discussion  of  the  Equation.  —  Intercepts  and  Extent.  — 

The  intercepts  of  equation  (^10)  are  obviously    ±  a  on  the 

a>axis,  and  ±  6  on  the  2/-axis. 

Solving  for  ?/,  we  have  y=±-  Va-  —  .i--,    a    form    which 

a 

shows  that  all  values  of  x  numerically  greater  than  a  must 


THE   ELLIPSE 


87 


be  excluded.  Similarly  all  values  of  y  numerically  greater 
than  h  must  be  excluded.  Hence  the  ellipse  lies  wholly 
within  the  rectangle  whose  sides  are  x  =  ±  a,  y  =  ±  b. 

Axes  of  Symmetry.  —  The  form  of  equation  (20)  shows 
that  the  ellipse  is  symmetrical  with  respect  to  both  the 
O/'-axis  and  the  y-axis,  and  hence  is  symmetrical  with  respect 
to  the  origin  0. 

The  segment  of  the  prin- 
cipal axis  intercepted  by 
the  ellipse,  namely  W,  is 
called  the  major  or  trans- 
verse axis.  The  segment 
BB'  of  the  2/-axis  is  called 
the  minor  or  conjugate  axis. 
The  major  axis  is  of  length 
2  a  and  the  minor  axis  of 
length  2  6.  0  is  called  the  center  of  the  ellipse  and  is 
the  mid-point  of  every  chord  of  the  ellipse  whicli  passes 
through  it,  since  it  is  the  center  of  symmetry. 

From  the  definition  of  b  (§  61),  we  have  at  once 


E 

Y 

E' 

X 

B 

--.^^ 

n' 

D 

V\    F          0 

M  F'JV 

D'X 

whence 


(21) 


a  relation  showing  that  a  is  always  greater  than  either  b  ore. 
As  equations  (18),  (19),  and  (21)  are  mutually  independent, 
all  of  the  five  constants  of  the  ellipse  can  be  found  if  any 
two  of  them  are  given. 


Exercise  2. 

minor  axis. 


Find  the  distance  from  the  focus  to  the  ends  of  the 


63.  Second  Focus  and  Directrix.  —  On  OV^  take  OF'  =  0F 
and  OB'  =  OD.  Draw  D'E'  parallel  to  DE.  Then  F'  is 
also  "a  focus  and  D'E'  the  corresponding  directrix  of  the 
ellipse.     This  is  shown  as  follows  : 


88 


ANALYTIC   GEOMETRY 


Let  P  be  any  point  on  the  ellipse.     Draw  F'P,  and  draw 
PN'  perpendicular  to  D'E'.     Calling  FP  =  e  •  PN\  and  sub- 
stituting the   corresponding 


values,  the  result  is 


V( 


ae 


xy  +  ?/'  =  e'  —  ^ 


Squaring  both  sides  and  col- 
lecting terms,   this   reduces 
to  equation  (20).     Hence  the 
ellipse  of  eccentricity  e  and 
having    D'E'    for    directrix 
and  F'  for  focus  is  the  same  as  that  referred  to  DE  and  F. 
Therefore  the  ellipse  has  two  foci,  (±  c,  0),  and  two  direc- 
trices, X  =  ± 


a 


0 


PROBLEMS 

l/'l.  Find  the  semi-major  axis,  tlie  semi-minor  axis,  the  eccentricity, 
the  coordinates  of  the  foci,  and  the  equations  of  the  directrices  of  the 
following  ellipses  :  (a)  9  x^  +  2b  y'^  =  225. 

(SoZw^io/i.  —  Dividing  by  225,  we  have 

25^9"    ' 
which  is  of  the  standard  form.     Hence  a  =  5,  and  b  =  3.     By  (21)  c  =  4 
and  by  (19)  e  =  |.     The  foci  are  (±  4,  0)  and  the  directrices  x  =±6  J. 

(&)  9^2  +  .36  ?/2  =  324  ;  V(e)  x^  +  2y^  =2; 

'-'■y  k<c)  5  x2  +  8  ?/2  =  40  ;  (/)  4  a:2  +  9  y2  3,  1  . 

-^^        (^)  4 a:2  +  9  2/2  =z  36  ;      ^'       (g)  9x'^  +  16y'^  =  16. 

2.    Write  the  equation  of  the  ellipse  with  the  center  at  the  origin 
and  the  principal  axis  as  the  a:-axis,  given  : 

(a)  semi-major  axis  5,  eccentricity  f  ; 
(6)  semi-minor  axis  4,  eccentricity  |  ; 
^c)  vertex  (13,  0),  focus  (5,  0)  ; 
(^d)  focus  (4,  0),  e  =  I; 
^  (e)  semi-major  axis  6,  directrix  x  =  3\/5  ; 
(/)  directrix  x  =  —  4,  eccentricity  |  ; 
(gr)  one  focus  ( ^2,  0),  one  directrix  x  =  2V2. 
^.  ^    ^  'f  Ans.    (a)  7  x2  -f-  16 1/2  =  175  ; 

(d)  3  x2  -}-  4  1/2  =  192  ; 
ig)  a:2  +  2  y^  =  4. 


THE   ELLIPSE  89 

/  /     r/o-    Express  a  in  terms  of  e  and  ».  Ans.  —^ — . 

^       4.    Express  c  in  terms  of  e  and  p. 

'  y  ^  5.   Find  the  locus  of  a  point  moving  so  that  the  sum  of  its  distances 
/"^from  the  points  (  ±  12,  0)  is  26. 

6.  Solve  Problem  5  when  the  points  are  (±  c,  0)  and  the  sum  of 
the  distances  is  2  a. 

7.  From  the  circumference  of  a  circle  of  radius  4  a  series  of  per- 
pendiculars is  drawn  to  a  diameter.  A  point  moves  so  as  to  bisect 
each  perpendicular.     Find  its  locus. 

8.  Generalize  Problem  7,  calling  the  radius  a  and  assuming  that 

the  point  cuts  off-th  part  of  each  perpendicular.  Ans.  x"^  +  ti^y^  =  a'^. 
n 

9.  Prove  that  the  projection  of  a  circle  upon  a  plane  is  an  ellipse. 

Hint.  —  Let  the  plane  upon  which  the  circle  is  projected  intersect  the 
plane  of  the  circle  in  a  diameter,  and  let  the  angle  between  the  two  planes 
be  a.    Then  a  is  the  radius  of  the  circle  and  h  =  a  cos  a. 

10.  Find  the  locus  of  the  middle  points  of  the  ordinates  of  the 
ellipse  h'^x^  +  d^y^  =  a^b'^. 

11.  Lines  are  drawn  from  the  center  of  an  ellipse  to  its  perimeter. 
Find  the  locus  of  a  point ,which  divides  these  lines  in  the  ratio  2  :  3. 

^  Ans.   25  &2x2  +  25  aV  =  4  a^b^. 

"^12.    Find  the  locus  of  the  vertex  of  a  triangle  whose  base  is  2  a  and 
such  that  the  product  of  the  tangents  of  its  base  angles  is  —  • 

64.  Focal  Radii.  —  The  distances  of  any  point  on  the 
ellipse  from  the  foci  are  called  the  focal  radii.  Denoting 
these  by  p  and  p',  we  have  from  §§61  and  63, 

p  z=FP  =ef-  +  x\=  a  +  ex, 
and  p' =F'P  =  ef--x\=a-ex. 


Adding  these,  p-{-p  =  2  a. 

Hence  the  sum  of  the  distances  of  any  point  on  the  ellipse 
from  the  foci  is  a  constant  and  equal  to  the  major  axis.     This 


90 


ANALYTIC   GEOMETRY 


in  connection  with  Problem  6,  page  89,  shows  that  the 
ellipse  may  be  defined  as  the  locus  of  a  point  which  moves 
so  that  the  sum  of  its  distances  from  two  fixed  points  is  a 
constant. 

65.  Latus  Rectum.  —  The  chord  through  either  focus  per- 
pendicular to  the  major  axis  is  called  the  latus  rectum.  If 
in  the  equation  of  the  ellipse  we  put  x  =  ±  c,  then 


4 +  ?-  =  !, 


?/-  _  a^  —  c^  _  h"^ 


a' 


62 


of  the  latus  rectum  is 


h 
a 


a' 


a' 


Therefore  ^  =  ±  -,  or2/=±— •     Hence  the  length 

ha  a 


a 


As  in  the  case  of  the  parabola,  the  location  of  the  ends 
of  the  latus  rectum  is  a  considerable  aid  to  accuracy  in 
sketching  an  ellipse  of  given  dimensions. 

Exercise  3.     Show  that  the  latus  rectum  is  also  given  by  2  ep. 


66.    Ellipse  with  Foci  on   F-axis.  —  If  the  foci  are  on  the 
?/-axis,  this  becomes  the  principal  axis.     Calling  the  major 

axis  and  minor  axis  of  the 
ellipse  2  a  and  2  h  respectively 
as  before,  the  effect  on  the  equa- 
tion of  the  ellipse  is  merely  to 
interchange  the  variables  x  and 
y.     It  is  then 


^  +  "^  =  1, 

or  hhj"^  +  cv^x"^  =  a^lP. 


(20  a) 


In  this  form  it  is  readily  seen 
that  the  coordinates  of  the  ver- 


THE   ELLIPSE  91 

tices  are  (0,  ±  a) ;   the  coordinates  of  the  foci  are  (0,  ±  c) ; 

the  equations  of  the  directrices  are  y  =  ±  — ;  and  the  lengths 
of  the  focal  radii  are  a  ±ey.  ^ 

Exercise  4.  Prove  as  in  §  61  that  the  central  equation  of  an  ellipse 
having  the  foci  on  the  y-axis  is  b^y^  +  a^x^  =  a^W. 

Exercise  6.  Prove  that  the  focal  radii  of  the  ellipse  with  foci  on 
the  y-axis  are  a  ±  ey. 

67.  The  Circle  as  a  Limiting  Form.  —  If  5  =  a,  the  equa- 
tion  of  the  ellipse  becomes  — \-  ^  =  1 ,  ot  x"^  +  y"^  =  a"^,  which 

is  a  circle  with  the  center  at  the  origin  and  radius  a.     The 

constant   c  equals  Va^  —  a"^  =  0,  and   hence  e  =  -  =  0,  and 

a 

-  is  infinite.  The  circle  then  may  be  considered  as  a  limit- 
ing form  of  the  ellipse  with  eccentricity  zero,  the  two  foci 
coinciding  at  the  center  and  the  directrices  removed  an 
infinite  distance  from  the  center. 

68.  Construction  of  an  Ellipse  by  Points.  —  When  a  careful 
drawing  of  an  ellipse  is  required,  it  is  always  possible  to 
form  the  equation  and  compute  a  table  of  values.  This, 
however,  entails  considerable  labor,  and  it  is  usually  better 
to  proceed  according  to  one  of  the  following  methods. 

1.  Given  the  directrix,  focus,  and  eccentricity.  Draw  a 
parallel  to  the  directrix  at  a  distance  k,  and  with  a  radius 
ek  and  the  focus  as  a  center  draw  an  arc.  Its  intersections 
with  the  parallel  are  two  points  of  the  ellipse.  By  varying 
k,  as  many  points  as  are  desired  may  be  obtained.  When 
coordinate  paper  is  used  the  parallels  are  already  drawn ; 
and  if  e  is  a  convenient  number,  the  points  can  be  readily 
located. 

2.  Given  the  foci  and  the  major  axis.  Lay  off  the  major 
axis  on  coordinate  paper.     Then  with  a  radius  p  <  2  a  and 


92  ANALYTIC    GEOMETRY 

a  focus  as  a  center  draw  an  arc.  With  the  other  focus  as  a 
center  and  /j'  =  2  a  —  p  as  a  radius,  draw  another  arc  cutting 
the  first  in  two  points.  These  are  points  of  the  ellipse  by 
§  64.     By  varying  p  we  get  as  many  points  as  are  desired. 


PROBLEMS 

y^-  ^1.  Find  the  vertices,  foci,  eccentricity,  equations  of  the  directrices, 
and  length  of  the  latus  rectum  of  the  ellipse  whose  equation  is  25  x^  + 
9  y'2  =  225. 

Solution.  —'D'lyiding  by  225,  we  have 

9      25 
Since  a  is  always  >  &,  «  =  5  and  6  =  3,  and  the  equation  is  of  the  form 

a2      62 
Hence  the  vertices  are  (0,  ±5),  the  foci  (0,  ±4),  e  =  |,  the  directrices  are 
?/  =±  6|,  and  the  latus  rectum  ^i-. 

-l  2.    Investigate  the  following  equations  as  in  Problem  1. 

/•  (a)  36a:2  +  4?/2  =  144;                          (d)     9  x^  +  ^y'^  =  4:-, 

YaA-  ^h)     3x2  +  21/2  =  1;                             (e)    16  a;2  +  9  ?/2  =  144 ; 

V  (c)      5  x2  +  3  ?/2  =  1  ;                             (/)   25  x^  +  4y^=  100. 


'{^ 


fi 


'    3.    Write  the  equation  of  the  ellipse  with  the  center  at  the  origin, 

given  : 

>{a)  focus  (0,  6),  directrix  y  =  S; 

.^(&)  vertex  (0,  4),  focus  (0,  3)  ; 

(c)  directrix  x  =  6,  distance  between  foci  7  ; 

(d)  latus  rectum  —,  vertex  (0,  —  13)  ; 

lo 

(e)  one  end  of  minor  axis  ( V2,  0),  latus  rectum  2. 

Ans.  (a)  4  x2  +  2/2  ^  48  ;  (^d)   169  x2  +  25  2/2  =  4225. 

My^.   Write  the  equations  of  the  following  ellipses  with  the  center 
<X  the  origin,  the  foci  being  first  on  the  x-axis  and  second  on  the  y-axis, 
unless  the  given  data  fix  their  position : 

(^)  a  =  8,  6  =  6  ;  (c)  c  =  3,  e  =  | ; 

^)   e  =  ip  =  3;  (d)  c  =  5,  5  =  12; 

^e)   vertices  (±  5,  0),  one  focus  (3,  0)  ; 

(/)    foci  (±  5,  0),  directrix  x  =  15  ; 
i'(^)   vertices  (±  13,  0),  e  —  xV- 


/^^A^,/.K<it    yfJL'f^JO^Jr^r^      "^ 


c^  X 


THE   ELLIPSE 


93 


5.  Construct  by  the  methods  of  §  68  ellipses  for  which 

(a)  ;)  =  24,  e  =  1 ;  (c)    a  =  20,  c  =  12  ; 

(5)  i?  =  16,  e  =  i  ;  (d)  a  =    8,  c  =    5. 

6.  Find  the  center,  vertices,  and  ends  of  the  minor  axis  of  the  fol- 
lowing ellipses  and  sketch  each  cui^^e. 

(a)  Focus  (6,  2),  directrix  x  =  12, 

Solution.  —  Here  p  =  6.  Using  for- 
mulas (18)  and  (19),  we  have  a  =  4 
and  c  =  2.  Hence  the  center  is  (4,  2) 
and  the  vertices  (8,  2)  and  (0,  2) .  By 
(21)  b  =  vl2  and  the  ends  of  the  minor 
axis  are  (4,  2±Vl2). 


(b)   Focus  (4,  0),  directrix  x  =  0, 


e  = 


—  2  • 

3  J 


OTH 


(c)  focus  (1,  2),  directrix  x  =  —  4,  e  =  h  ; 

(d)  focus  (3,  —  4),  directrix  ?/  =  0,  e  =  | ; 

(e)  focus  (—  2,  3),  directrix  y  =  —  1,  e  =  ^. 

Derive  the  equations  of  the  ellipses  of  Problem  6. 

Ans.  (a)  3  X'-  —  24  a:  -I-  4  2^2  —  16  y  -I-  16  =  0. 
8.   A  line  of  constant  length  moves  so  that  its  extremities  are  on 
two  lines  at  right  angles.     Find  the  locus  of  a  point  P  on  the  line  at 
a  dist^ce  a  from  one  end  and  b  from  the  other. 

/^.    Find  the  equation  of  an  ellipse  having  its  center  at  the  origin 
and  axes  along  the  coordinate  axes,  which  passes  through  : 

(a)   (2,  4)  and  (5,  -3)  ;  (b)   (1,  2)  and  (2,  1). 

10.   Prove  that  the  latus  rectum  of  an  ellipse  is  a  third  proportional 
to  the  major  and  minor  axes. 

69.  Mechanical  Construction  of  the  Ellipse.  —  The  relation 
of  the  focal  radii  may  be  used  to  construct  the  ellipse  me- 
chanically. On  a  drawing  board 
fasten  two  tacks  at  the  foci  F 
and  F',  and  tie  a  string  about 
them  of  length  equal  to  2  a  -f  2  c. 
^  If   a   pencil    is    placed    in    the 

^^-.^^         ^y  loop    FPF'    and    moved    so    as 

to  keep  the  string  taut,  then 
PF  +  PF'  is  a  constant  equal  to  2  a,  and  P  describes  an 
ellipse  (§  64). 


94 


ANALYTIC   GEOMETRY 


70.  The  Ellipse  as  a  Section  of  a  Cone.  —  Let  V-AEBD 
be  a  right  circular  cone,  and  VAB  a  plane  determined  by 
the  vertex  V  and  the  diameter  of  the  base  AB.     This  plane 

will  be  perpendicular  to 
the  base  and  will  con- 
tain the  elements  VA 
and  VB,  and  also  the 
axis  of  the  cone  and  the 
center  of  every  section 
parallel  to  the  base. 

Pass  a  plane  perpen- 
dicular to  the  plane 
VAB,  intersecting  VA 
and  VB  at  M  and  N. 
Through  C,  the  mid- 
point, and  Q,  any  other 
point  of  3/JV,  pass  planes 
parallel  to  the  base, 
intersecting  the  plane 
VAB  in  FG  and  IIK; 
Zhese  sections  are  circles  and  FG  and  HK  are  their 
diameters. 

QP,  the  intersection  of  the  planes  FPG  and  3IPX,  is 
perpendicular  to  the  plane  VAB,  and  hence  it  is  also  per- 
pendicular to  the  lines  FG  and  MN.  In  the  circles  FPG 
and  HRK, 

QP'=FQ'  QG    and     CM''  =  HC  -  CK. 

The   lines   FQ  and   IIC  are   parallel,  hence  the   triangles 
FQM  and  HCM  are  similar. 

,   Z2  =  ^andS!?  =  M. 
HC     MC  CK     CN 


Hence 


QI^_MQ  ,  QN^ 


CH 


2     MC      CN 


(a) 


THE   ELLIPSE  95 

Let  CR  =  b  and  3IC  =  CN  =  a,  and  we  have 

QP'_3IQ^QX 


a- 


(6) 


Xow  as  Q  moves  along  3fX,  P  moves  along  the  mtersec- 
tion  of  the  conical  surface  and  the  cutting  plane.  Calling 
MN  the  .T-axis  and  CH  the  ^/-axis,  the  coordinates  of  P  will 
be  X  =  CQ  and  y  =  QP.  Therefore  MQ  =  a-\-x  and  QX= 
a  —  X.     Substituting  in  (li),  we  have 

y^_  (a  4-  x){a  —  x) 
b^-~  a2 

which  reduces  to 

o?     IP- 
Thus  the  section  of  the  cone  is  an  ellipse. 

Exercise  6.  Show  that  the  section  of  a  right  circular  cylinder  made 
by  a  plane  not  parallel  to  the  base  is  an  ellipse. 


CHAPTER   VI 


THE   HYPERBOLA 

71.    Constants.  —  In  §  52  the  hyperbola  was  defined  as  a 
conic  of  eccentricity  greater  than  1. 

In  the  figure  let  AE  be  the  directrix,  F  the  focus,  and  3/JY 

the  principal  axis.  Let  the 
distance  between  D  and  F 
be  denoted  by  p,  and  the 
eccentricity  by  e. 

Since  e  >  1,  there  are  two 
points  of  the  hyperbola,  V 
and  F',  on  the  principal 
axis,  one  on  each  side  of  the 
directrix.  As  in  the  ellipse 
these  are  called  vertices; 
the  point  C  midway  between 
them  is  called  the  center ; 
and  the  distances  between  V  and  C,  and  F  and  C  are  des- 
ignated by  a  and  c  respectively. 

Applying  the  definition  of  the  hyperbola  to  the  points  V 
and  V\  we  have 


and 


VF=  e  •  DV  or  c—  a  =  e[a  —(c—p)'], 
y'F=  e  .  V'D  or  c -\- a  =  e^a -\-  (c  —  jj)]. 


Subtracting  and  adding  these  equations, 


a  =  e(c—2J)  or  c  -  p 


a 


and 


e 

c=  ae. 


(22) 
(23) 


96 


THE   HYPERBOLA 


97 


The  first  relation  gives  the  distance  between  the  center 
and  directrix  in  terms  of  a  and  e,  and  the  second  that  be- 
tween the  center  and  focus. 

Exercise  1.  From  the  definition  of  the  hyperbola  find  D  V  smdV'D 
in  terms  of  p  and  e.  .  p 

e±l 

Exercise  2.  Using  the  results  of  Exercise  1,  show  that  C  is  to  the 
left  of  the  directrix. 


72.  The  Equation  of  the  Hyperbola.  —  To  derive  the  equa- 
tion of  the  hyperbola  take  the  principal  axis  as  the  a>axis  and 
the  center  as  the  origin. 
Then  F  has  the  coordinates 
(ae,  0),   and   the   directrix 

has  the  equation  x  =  -- 

e 

Let  P  be  any  point  on 
the  locus.  Draw  PiV  per- 
pendicular to  AE  and  join 
FF.     Then 


FF=e  .  NF. 


But 


FF  =  V{x  -  aef  -f  2/^  and  NF  =  x 


a 


Hence  ^(x  —  aey-\-y^  =  elx  —  -\ 

Squaring  and  collecting  terms, 

(1  -  e2)a;2  -f  2/2  =  a\l  -  e"), 


or 


t 


a2(l  -  e2) 


=  1. 


In  this  form  the  equation  of  the  hyperbola  appears  identi- 
cal with  that  of  the  ellipse  (§  61).  However,  e  >  1  and 
a2(l  —  e^)  is  negative.  Calling  IP-  =  (^{^^  —  1),  we  have  the 
hyperbola  equation  in  its  usual  form,  when  the  center  is 


98  ANALYTIC    GEOMETRY 

taken  as   the  origin  and  the  principal  axis  as  the  x-axis, 
namely, 

---=1,  (24) 

or  61^2  —  ahf  =  ci^W. 

73.    Discussion  of  the  Equation.  —  Intercepts  and  Extent.  — 

Solving  equation  (24)  for  x,  we  have  x=  ±- ^if  +  5^,  which 

h 

shows  that  the  .T-intercepts  are  ±  a,  and  that  no  value  of  y 

needs  to  be  excluded. 


Solving  for  y,  y  =z  ±-  -y/x^  —  a"^,  showing  that  there  are 

a 

uo  y-intercepts  and  also  that  y  is  imaginary  for  all  values  of 

X  between  ±  a,  but  if  x  is  numerically  greater  than  a,  y  is 

real  and  increases  indefinitely  as  |  x  |  increases.     Thus  the 

hyperbola  consists  of  two  infinite  branches,  lying  without 

the  lines  x  =  ±  a. 

Symmetry.  —  The  form  of  (24)  shows  that  the  hyperbola  is 
symmetrical  to  the  x-  and  ?/-axes  and  hence  to  the  origin. 
The  segment  V'Voi  the  principal  axis  is  called  the  transverse 
axis  and  is  of  length  2  a.  The  segment  B'B  of  the  ?/-axis, 
where  B'O  =  OB  =  b,  is  called  the  conjugate  axis.  From  the 
definition  of  6  in  §  72  we  have  at  once 

c-  =  a2  +  &2^  (25) 

a  form  showing  that  for  the  hy2:>erlx)la  c  is  greater  than 
either  a  or  6,  while  there  is  no  restriction  on  the  relative 
sizes  of  a  and  h.  It  is  for  this  reason  that  the  terms  major 
and  minor  axis  are  not  used. 

74.  Second  Focus  and  Directrix.  —  As  in  the  case  of  the 
ellipse  there  is  a  second  focus  and  a  second  directrix.  For 
on  the  principal  axis  take  OF'  =  OF  and  0Z>'  =  OD,  and 


THE   HYPERBOLA 


99 


erect  the  perpendicular  D'E'.  Then  the  same  curve  will  be 
generated  by  usmg  the  focus  F',  the  directrix  D'E',  and 
the  same  value  for  e. 

For  we  have  F'P  =  ey'P- 
but    F'P  =  V(ae  +  if  +  y\ 

and  .Y'P  =  .r  +  ^^     Therefore 

and  this  reduces  to  (24). 

Hence  the  hyperbola  has 
two  foci,   (±c,  0),  and  two 

directrices,  x  =  ±-' 


75.  Latus  Rectum.  —  The  chord  L'L  through  either  focus 
perpendicular  to  the  principal  axis  is  called  the  latus  rectum 
and  it  is  seen  (as  in  the  case  of  the  ellipse)  that  its  length- 

2b'^ 


IS 


a 


Exercise  3.     Show  that  the  length  of  the  latus  rectum  is  also  2  ep. 


PROBLEMS 

1.  Find  the  semi-transverse  au{ 
semi-conjugate  axes,  the  coordinates  ol 
the  foci,  the  eccentricity,  and  the  equa- 
tions of  the  directrices  of  the  hyperbola 
having  the  equation  25  x-  —  9  y-  =  225. 

Solution.  —  Dividing  by  225,  we  have 
the  standard  form 

9       25 
Hence    « =  3,    6  =  5,     c=  VsJ,     and 
e  =  iV^.     The  foci  are  (±  V34,0),  and 

9 


the  directrices  are  a:  =  ± 


Vsi 


As  the 


we  lay 
points. 


vertices    (zb  3,  0)  give  only  two  points, 
off  the  semi-latus  rectum —  =  ^  from  the  foci,  getting  four  more 


100 


ANALYTIC    GEOMETRY 


/  2.   Find  the  properties  of  the  following  hyperbolas  as  in  Problem  1 : 
(a)     9x--^-16y2=  144;  (d)       x^  -    2y'  =  l; 

VV\>  ^(&)  25x2  _    9^-2  =  225  ;  "(e)    3x2  -    5  2/2  ^  15  . 

/^'X       ^'^        ""'"    9^2  ^9.  (/)  25x2 -15  2/2  =  144. 

^  ^-:.3-    i'i"d  the  equation  of  the  hyperbola  having  its  center  at  the 
k\  rajaigin  and  its  foci  on  the  x-axis,  if 

y(a)  a  =  4,  6  =-; 


(d)  a  =  5, 

(e)  c  =  5, 


e  = 


3' 

a=2\/G; 


^  (c)  6  =  4,  e  =  2  ; 


)((/)  c  =  V5,  e  = 


V5 


^, 


^?iS. 


ep 


e2-l 


y^4.   Express  a  in  terms  of  e  and  p. 

5.  Express  c  in  terms  of  e  and  p. 

6.  Find  the  locus  of  a  point  moving  so  that  the  difference  of  its 
distances  from  the  points  (i  13,  0)  is  24. 

A71S.   25  x2  -  144  ?/2  =  3600. 

7.  Solve  Problem  6  when  the  points  are  ( ±  c,  0)  and  the  difference 
of  the  distances  is  2  a. 

8.  Define  the  hyperbola  from  the  conditions  of  Problem  7. 

^  9.    Find  the  equation  of  a  hyperbola  having  its  center  at  the  origin 
\iE«id  its  foci  on  the  x-axis  and  passing  through  : 

(a)   (3,  2)  and  (-  \/3,  1)  ;         (ft)   (3,  0)  and  (5,  -  4). 
10.   Find  the  equation  of  a  hyperbola  whose  vertex  is  midway  be- 
tween the  center  and  the  focus.  Ans.   Sx^  —  y-  =  S  a?-. 


76.  Hyperbola  with  Foci  on  the  F-Axis.  —  If  we  describe  a 
hyperbola  with  the  foci  on  the  i/-axis,  with  its  transverse 
axis  2  a,  distance  between  the  foci  2  c,  and  eccentricity  e, 

the  effect  on  the  original 
equation  will  be  to  inter- 
change the  variables  x  and 
2/,  and  the  equation  will  be- 
come 

a2      IP- 

or      hhf  —  ci?x^  =  (J^IP-. 

The  vertices  of  this  hy- 


F 

V 

r^ 

P 

D 

1 

0 

Q            X 

U'                       ,' 

E' 

THE   HYPERBOLA 


101 


perbola   are    (0,    ±  a),   the   foci    (0,    ±  c),   and   the   direc- 


trices 


,  a 
e 


Exercise  4.     Derive  equation  (24a)  as  in  §  72. 

77.   Focal  Radii.  —  As  in  the  ellipse  the  distances  of  any 

point  on  the  hyperbola  from  the  foci  are  called  the  focal 

radii  and  designated  by  p  and  p.     From  the  ligure  of  §  74 

we  have  / 

p  =  FP=  e'NP=elx--]=ex-  a, 


and 


p'  =  F'P  =  e.N'P  =  e(x-^-]=ex-\-a. 


Subtracting  these,  we  have  p  —  p  =  2  a.  Hence  the  dif- 
ference of  the  distances  of  ciny  point  of  the  hyperbola  from  the 
foci  is  constant  and  equcd  to  the  transverse  axis.  This  agrees 
with  the  results  of  Problems  7  and  8,  page  100,  which  show 
that  the  hyperbola  may  be  defined  as  the  locus  of  a  point 
moving  so  that  the  difference  of  its  distances  from  two 
fixed  points  is  constant. 

78.  Asymptotes.  —  If  a  line  is  drawn  through  the  center 
of  a  hyperbola  intersecting  it  at  P,  and  P  is  made  to  move 
off  to  infinity  along  the 
curve,  the  line  will  turn 
about  0  and  approach  one 
of  two  limiting  lines.  We 
call  these  asymptotes,  defin- 
ing them  as  lines  through 
the  center  of  the  hyperbola 
along  which  the  curve  re- 
cedes to  infinity. 

The  equation  of  any  line  PP^  through  the  center  or  origin 
is  ?/  =  7nx.  The  coordinates  of  its  intersections  with  the 
hyperbola  are   given   by  solvmg  this  simultaneously  with 


\ 

\ 

Y 

y^^p 

/ 

J' 

;?o 

X 

^ 

V 

102  ANALYTIC   GEOMETRY 


-^  =  1.      Substituting,    we    have =  1,    which 

gives  ^^ 

x  =  ± 


Now  if  P  is  to  move  off  to  infinity,  x  must  become  in- 
finite, i.e.  the  denominator  must  become  zero.     This  gives 

6^  —  ahn~  =0     or     m  =  ±  - . 

a 

Hence  the  slopes  of  the  asymptotes  are   ±  -    and   their 

a 

equations 

y  =  ±lx,  (26) 

a 

or  bx  ±  ay  =  0. 

These  equations  may  be  combined  by  §  42  in  the  form 

b^x^-aY=0,     or    ^-^1  =  0.  (26a) 

a^      b~ 

In  case  the  vertices  are  on  the  ?/-axis,  x  and  y  are  inter- 

changed,  giving  -- —  —  =  0.     In  either  case,  to  find  the  equa- 
o}      b'^ 

tions  of  the  asymptotes,  ivrite  the  equation  of  the  hyperbola  in 

standard  form,  replace  the  1  by  0,  and  factor. 

79.  Conjugate  Hyperbolas.  —  Since  the  relative  sizes  of  a 
and  b  are  immaterial,  a  hyperbola  may  be  described  having 
the  segment  2  b  for  a  transverse  axis  and  the  segment  2  a 
for  a  conjugate  axis.  The  hyperbola  with  transverse  axis 
A' A  (=  2  a)  along  the  a^axis  and  conjugate  axis  B'B  (=  2  b) 

/v»2  nil 

along  the  ?/-axis  has  the  equation ^=  1.     The  hyper- 
bola with  transverse  axis  B'B  (=2  6)  along  the  ^/-axis  and 


THE   HYPERBOLA 


103 


conjugate  axis  A' A  (=  2  a)  along  the  ic-axis  has  the  equation 


y~      X- 


=1.     The  two  hyperbolas 


(«)  ^-!^=i  =»"'^  (*)  ?!-^!=i 


are  called  conjugate  hyperbolas.     The  transverse  axis  of  each 
is  the  conjugate  axis  of  the  other. 


Since  c^  =  a-  +  6-,  the  foci  of  both  hyperbolas  are  at  the 
same  distance  from  the  center.  Those  of  (a)  are  ( ±  c,  0)  ; 
those  of  (h)  are  (0,  ±  c).  If  we  let  e^  and  e,  be  the  respec- 
tive eccentricities,  we  have  ei=  -,  eo=  -,  whence  e^ :  62= &  :  a. 

a     '     h 

The  directrices  of  (a)  are  x  =  ±  -,  oi  (b)  y  =  ±—.     From 

the  equations  it  is  evident  that  conjugate  hyperbolas  have 
the  same  asymptotes. 

This  property  is  useful  in  sketching  a  pair  of  conjugate 
hyperbolas.  Draw  the  rectangle  of  the  given  axes,  that  is, 
a  rectangle  having  its  sides  of  length  2  a  and  2  b  and  sym- 


104 


ANALYTIC   GEOMETRY 


metrical  with  respect  to  the  axes,  and  draw  the  diagonals.    As 

these  have  slopes  ±  - ,  they  are  the  asymptotes.     The  circle 

a  

circumscribing   the  rectangle  has    the  radius  V'a^  -\-b~  =  c, 

and  therefore  intersects  the  axes  at  the  foci  of  the  required 

hyperbolas.     AVith  the  rectangle  and  asymptotes  as  guiding 

lines  the  curves  can  be  sketched  accurately. 

Exercise  5.     Prove  that  the  directrices  of  a  hyperbola  cut  the  con- 
jugate hyperbola  at  points  on  the  circle  drawn  through  the  four  foci. 


When  a  =  b,  the   equation 


i/  =  a- 


80.  Equilateral  Hyperbolas. 

of  the  hyperbola  reduces  to 

and  tlmt  of  the  conjugate  to 

7/2  _  3^2  _  ^2^ 

The  asymptotes  are  inclined  to  the  axes  at  45°  and  are  per- 
pendicular to  each  other.  Such  hyperbolas  are  called  equi- 
lateral or  rectangular  hyperbolas. 

81.  Construction  of  a  Hyperbola  by  Points.  —  A  hyperbola 
may  be  plotted  accurately  by  forming  a  table  of  values 
from  the  equation  or  by  proceeding  according  to  methods 
similar  to  those  used  for  the  ellipse  (§  68). 

PROBLEMS 

1.  Find  the  semi-transverse  and  semi- 
conjugate  axes,  the  coordinates  of  the 
foci,  the  eccentricity,  and  the  equations 
of  the  directrices  of  the  hyperbola  whose 
equation  is  25  x-  -  ^  y^  -{-  225  =  0. 

)SoZ?<i?07i. —Transposing  and  dividing  by 
—  225,  we  have 

9      25 

Since  the  term  in  y  is  positive,  this  is  of 


the  form 


a2      62 


Hence  a  =  5,  6  =  3,  c  =  V34,  and  e  =  W'Sk. 


THE   HYPERBOLA  105 

,—  25 

The  foci  are   (0,,  ±  v34)   and   the  directrices  are   ?/=±— -z=      Diawing 

the  asymptotes,  the  curve  is  readily  sketched.  v  34 

i^  2.    Find  the  properties  of  the  following  hyperbolas  in  the  same 
K      manner  as  in  Problem  1  : 

/,               (a)        9x^-y^  =  lS;            ^)     Sx^  -  4:y^  -  12  =  0 ;  Qy2  ^ 

1/^^         ^6)   5  2/2  -  10 a:^  =  50  ;               (e)   9x2  -  18  2/- +  16  =  0  ;  'P^. 

/fc)        x2-9y23z9;                 (/)           16x2-25?/2  =  64.  t^^^V 

KZ.    Write  the  equations   of  the   hyperbolas  conjugate  to  thos^'of' 
i/^   Problem  2  and  of  their  common  asymptotes. 

4.   Find  the  equations  of  the  hyperbolas  of  Problem  3,  page  100,  if  the 
foci  are  taken  on  the  y-axis.  Ans.     (a)   64  x'-^  —  25 1/^  +  400  =  0. 

t/^fi.   Construct  by  points  hyperbolas  for  which 

(a)p  =  24,  e  =  3;  *^c)  a  =  20,  c  =  32; 

(5)  i)  =  16,  e  =  2 ;     ^f ^       ^{d)  a  =  5,  c  =  8. 

/6.    Find  the  center  and  lengths  of  the  semi-axes  of  each  of  the  fol- 
lowing hyperbolas.     Draw  the  asymptotes  and  sketch  each  curve. 

■  \a)  Focus  (6,  2),  directrix  a:  =  12,  e  =  2  ; 

^j'j^  Ans.     c  =8,  e=(14,  2),  a  =4,  6=4  V3. 

'         ^(&)  focus  (4,  0),  directrix  x  =  0,  e  =  \\ 

>J    (c)   focus  (1,  2),  directrix  a;  =  —  4,  e  =  2; 


(d)  focus  (3,  —4),  directrix  1/ =  0,  e  = 


_   4 


3  > 


(e)  focus  (—2,  3),  directrix  2/  =—1,  e  =  f . 

^  7.   Derive  the  equations  of  the  hyperbolas  of  Problem  6 
y^  /  z^..  ^^*-     («)     3  ic2  -  84  X  -  2/2  +  4  ?/  +  536  =  0. 

'8.    Show  that  the  distance  of  any  point  on  an  equilateral  hyperbola 
from  the  center  is  a  mean  proportional  between  the  focal  radii. 
9.    Show  that  for  a  pair  of  conjugate  hyperbolas  e^  +  ei^  —  e^e'?. 

82.  Mechanical  Construction  of  the  Hyperbola.  —  Place 
thumb  tacks  in  a  drawing  board  at  F  and  F' ^  the  foci.  Let 
a  string  be  tied  to  a  j)encil 
at  P  and  looped  about  the 
tacks  as  in  the  figure.  If 
the  ends  are  drawn  in  to- 
gether, P  will  describe  a 
hyperbola  since 

{F'F+FP)-F'P  ^'  \ 

remains  constant.     For  F'F  is  constant ;  hence  F'P  —  FP 
is  a  constant. 


106 


ANALYTIC   GEOMETRY 


83.  The  Hyperbola  as  a  Conic  Section.  —  Let  the  cutting 
plane  intersect  the  given  cone  so  as  to  meet  some  of  the 
elements  produced.     Follow  the   construction  of  §  70  and 

choose  the  axes  in  the  same  man- 
ner, viz.,  origin  at  C,  and  a; -axis 
3IX.     Then 


Also 

FQ^MQ     ^^^^     QG^m, 
KC     MC  HC     CN 

Substituting, 

^      MC'CN  ^ 

or  if 
MC  =  CN=a  and  KC-IIC  =  h\ 


¥ 


Vf Z)^        f  =  ^aiQ'NQ). 


But 
whence 


3IQ  =  X  -{-  a,     NQ  =  x  —  a, 


^       a^-  ^  (£-      If- 


Hence  the  plane  cuts  the  surface  of  the  cone  in  one  branch 
of  a  hyperbola  and  the  conical  surface  formed  by  prolonging 
the  elements  through  the  vertex  in  the  other  branch. 


CHAPTER   VII 

TRANSFORMATION  OF   COORDINATES   AND 
SIMPLIFICATION  OF  EQUATIONS 

84.  Change  of  Axes.  —  The  position  of  the  axes  of  coordi- 
nates to  which  a  given  locus  is  referred  is  arbitrary.  It 
may  be  changed  at  will  and  the  equation  of  the  locus  altered 
to  correspond  by  substituting  for  the  former  coordinates 
their  values  in  terms  of  coordinates  measured  from  the  new 
axes. 

The  advantage  to  be  secured  by  a  change  of  axes  is 
usually  a  simplification  of  the  equation,  and  this  is  best 
secured  by  choosing  a  position  of  symmetry  when  possible. 

When  the  new  axes  are  drawn  parallel  to  the  old,  tli6 
transformation  of  the  equation  is  called  a  transformation 
by  translation.  When  they  are  drawn  through  the  same 
origin  oblique  to  the  axes  but  still  perpendicular  to  each 
other,  the  transformation  is  called  a  transformation  by 
rotation. 


85.  Formulas  of  Transla- 
tion.—  In  the  figure  let  the 
original  set  of  axes  be  OX 
and  0Y\  the  new  axes  O'X' 
and  0'  Y'  with  the  origin  at 
0',  which  has  the  coordi- 
nates (li,  Jc)  with  reference 
to  OX  and  OY.  Let  P  be 
any  point  in  the  plane  which 
has  coordinates  (x,  y)  with 
reference  to  the  old  axes,  and  (x^,  y')  with  reference  to  the 
new  ones.     Then 

107 


1 
Y 

Y 

P(x,y) 

D 

E 

1  {^\u) 

O' 

{h,k) 

B\          X' 

C 

0 

A         X 

108 


ANALYTIC   GEOMETRY 


(27) 


x  =  EP,         a/  =  DP,         h=--OC  =  ED, 
y  =  AP,        y'  =  BP,        k  =  CO'  =  AB. 

From  the  figure 

EP  =  DP  -  DE  =  DP  -\-  ED  and  AP=  BP  +  AB. 

Hence  x  =  x'  +  h, 

and  y  =  y'  +  k. 

These  are  the  formulas  of  translation. 

By  using  these  substitutions  a  given  equation  is  trans- 
formed into  a  new  equation  in  x'  and  y' ,  which  is  the  equa- 
tion of  the  locus  of  the  given  equation  with  respect  to 
coordinate  axes  drawn  through  the  point  (Ji,  k)  parallel  to 
the  old  axes. 

86.  Formulas  of  Rotation.  —  Let  the  original  axes  be  OX 
and  OY,  the  new  axes  OX'  and  OY',  and  the  angle  of  rota- 
tion 6.  P  is  any  point  in 
the  plane  with  coordinates 
(x,  y)  and  (xf,  y')  with  refer- 
ence to  the  old  and  new 
axes  respectively.     Then 

X  =0A,        xf  =  OD, 
y  =  AP,        y'  =  DP. 

Since  the  sides  of  angle 
CPD  are  perpendicular  to 
those  of  ^,  OPi)  =  ^.  Hence 
we  have 

CD  =  y  sin  (9,    CP  =  y' cos  0,    OB  =  x' cos  0,    BD  =  x'smO. 

But      x=OA=OB-  CD  and  y  =  AP=BD+  CP. 

Substituting,  we  get 

X  =  x'  cos  0  —  y'  sin  6,  (28) 

1/  =  x'  sin  6  +  y'  cos  6, 

which  are  the  formulas  of  rotation. 


TRANSFORMATION   OF   COORDINATES        109 

These  substitutions  transform  a  given  equation  into  a 
new  equation  with  variables  x'  and  ?/',  in  which  the  coordi- 
nate axes  are  drawn  through  the  old  origin  but  inclined  at 
an  angle  Q  to  the  old  axes. 

PROBLEMS 

General  Directions.  —  Usually  the  equation  obtained  by  transforma- 
tion is  a  known  form  and  its  locus  can  be  identified.  When  this  is  not  the 
case,  the  table  of  values  should  be  computed  from  the  new  equation  and 
the  curve  plotted  on  the  new  axes,  since  they  are  axes  of  symmetry.  Fre- 
quently the  intercepts  with  respect  to  both  old  and  new  axes,  together 
with  considerations  of  symmetry,  make  an  extended  table  of  values  un- 
necessary. 


V  >4ua 


Translate  the  origin  to  the  point  indicated  and  transform  the 
,^       quation  to  correspond.     Draw  the  curve  and  both  sets  of  axes. 

y^  (a)  4x2-?/-^ -8  X +  4?/ +  4  =  0,  (1,  2). 

A)i8.  4x'"2—  ?/'2  -f-4  =  0. 
^(6)    2x2 +  2/- 5  =  0,  (0,  .5); 

(c)    2/2  _  8^^  ^_  24,  (.3,0);  n        r 

^(d)   x2  4-  2/-  =  6x  -  4  2/,  (3,  -  2)  ;  r  :^.t^  \    . 

(e)    2/  =  ^'  +  6x2+8x  +  3,  (-2,  3); 
^v(/)  y  =  b  +  (x-a)%  (a,b); 
y^ig)    10x2  -  64x  +  9y2  4-  18  2/  -  71  =  0,  (2,  -  1). 

K  2.  Rotate  the  axes  through  the  angle  indicated  and  transform  the 
equation  to  correspond.  Draw  the  curve  and  both  sets  of  axes. 
Identify  known  loci. 

(a)  xy  =  8,d=^; 
^b)  x2  -  2/2  =  16,  0  =  ^  ; 
(c)  xy=-S,d  =  ~; 

^  (d)  x2  +  2/2  -  4x2/  +  9  =  0,  ^  =  ^; 

/(e)  x2  +  4x2/  +  2/2=  16,  ^  =  ^; 

^'(/)  29  x2  -  24  X2/  +  36  2/2  =  180,  ^  =  sin  -i  |  ; 
(g)  17  x2  -  312  xy  +  108  y^  =  900,  d  =  sin  -i  f . 

Ans.  («)  x'2  -  2/'2  =  16  ;  (/)  4  x'2  +  9  2/'  2  =  36. 


;^3.    Prove   that  x^  -\-  y^  =  r^  is   unchanged   by   rotating   the    axes 
through  any  angle. 


110 


/ 


,^(1 


ANALYTIC   GEOMETRY 


t   4.   Find  the  coordinates  of  P(4,  3)  and  P(2,  —1)  when  the  origin 
is  translated  to 

(«)   (6,  5)  ;  {h)  (4,  -  2)  ;  (c)  (-  2,  -  1)  ,  {d)  (-  4,  -  5). 
^^^  JL.  ^'is-  («)  (-2,-2),  (-4,  -6). 

^  5.    Find  the  coordinates  of 

(a)   (3,  5)  ;  (6)  (-  2,  2)  ;  (c)  (4,  -  2) ;  (cZ)  (-  6,  6) 
when  the  axes  are  rotated  45°;   —  46°;  90°. 

Ans.  (a)  (4V2,  V2),  (-  \/2,  4V2),  (5,  -  3). 


87.   Application  to  the  Conies.  —  Let  us  derive  the  equation 
of  the  conic,  taking  the  directrix  as  the  y-Sixis  and  the  prin- 
cipal axis  as  the  a;-axis.     AVe  have 
by  definition 

FP=eNP, 


,P{x,y) 


whence  ^(x—  py-^y-  =  ex, 

or  {x-py-\-y'  =  e''x\    (29) 


np,o) 


This  is  a  general  form  of  the 
conic  equation,  true  for  all  values 
of  e.  From  it  by  translation  of  axes  we  can  derive  the 
forms  previously  obtained. 

For  the  parabola  e  =  1.     Then  (29)  reduces  to 

2/^  =  2  2^x  —  p^. 

As  the  vertex  is  halfway  between  the  directrix  and  the  focus, 

its  coordinates  are  (^,  Oj.     Thus  to  move  the  origin  to  the 

vertex  we  substitute  cc  =  x'  -}-^-,  y  =  y' . 

Simplifying  and  dropping  primes,  we  get 

2/2  =  2  px, 

the  standard  equation  of  the  parabola. 

For  the  ellipse  we  have  seen  in  Chapter  V  that  the  dis- 
tance from  the  directrix  to  the  center  is  j)  +  c.     Therefore 


TRANSFORMATIOX  OF  COORDINATES    HI 

the  coordinates  of  the  new  origin  are  {^p  +  c,  0).     Substi- 
tuting x  =  a;'  +  j>  +  c,  ?/  =  ?/'  in  (29),  we  get 

(a;'4-c)2  +  2/'2  =  eV+i>  +  c)2. 

But  c^ae^p  -\-  c=-.     Using  these  relations  and  dropping 

primes,  we  obtain  the  equation,  ^1 

(x  +  aey  +  2/2  =  e'^fx  +  ^Y,  "  ^J,  ^ 

which  can  easily  be  reduced  to 

(l-e2)^.2^2/2  =  a2(l-e2). 

Dividing  by  the  right-hand  member,  and  putting  a2(l  —  e^)  =  b^, 
we  have  the  standard  form 

The  central  equation  of  the  hyperbola  can  be  derived  in 
the  same  way.  The  student  should  work  out  the  reduction 
for  himself,  bearing  in  mind  that  for  the  hyperbola  the  dis- 
tance from  the  directrix  to  the  center  is  j:>  —  c  =  —  - ,  and  that 
52  =  a2(e2-l).  ^ 

88.  Test  for  Axes  of  Symmetry.  —  To  find  axes  of  sym- 
metry parallel  to  the  coordinate  axes,  proceed  as  follows: 
Solve  the  equation  for  y  in  terms  of  x.  If  the  solution  is 
of  the  form  y  =  Z:  ±f(x),  the  line  y  =  k  is  an  axis  of  sym- 
metry. This  is  due  to  the  fact  that  each  value  of  x,  as 
X  =  a,  gives  two  values  of  y,  and  thus  two  points,  one  f(a) 
above  the  line  y  =  Jc,  the  other  f(a)  below.  Similarly,  if 
the  solution  for  x  is  of  the  form  x  =  h  ±  y(y),  the  line  x=h 
is  an  axis  of  symmetry.  Since  the  lines  x  =  h  and  y  =  Jc 
are  perpendicular  to  each  other,  a  curve  symuietrical  with 
respect  to  both  of  these  lines  has  their  intersection  (h,  k)  as 
a  center  of  symmetry.     (See  §  22.) 


112 


ANALYTIC    GEOMETRY 


89.    Simplification  of  Equations  by  Translation.  —  In  case 

the  curve  has  two  axes  of  symmetry,  they  can  be  found  by 
the  method  just  given  and  the  translation  made  by  formula. 
When  this  is  not  true,  the  second  method  illustrated  in  the 
following  examples  must  be  used. 

Example  1.  —  Simplify  x'^  +  4  y^  -  2  x  -  24:  y  -\-  21  =  0. 

Solution.  —  First  Method.     Solving  for 
X  aud  y,  we  have 

x  =  l±ny)* 
and  y  =  ?>±fj{x). 

Hence  the  axes  of  symmetry  are 

x=  \  and  ?/  =  3 
and  the  center  of  symmetry  is'(l,  3). 

Substitute 

X  =  x'  +  1  and  y  =  y'  -\-Z 
in  the  original  equation,  and  we  have 
(x'  +  1)2+4  {y'  +  3)2  -  2  {X'  +  1)  -  24  (^/'  +  3)  +  21  =  0. 
x'-^  +  \y'-^  =  16. 


16 


4 


Thus  the  curve  is  ap  ellipse  with  its  center  at  (1,  3)  and  semi-axes  4  and  2. 


Second    Method.     In    the    given    equation    substitute   x  — -  x'  +  h   and 
y  =  y'  -\-  k.    It  becomes,  after  collecting  coefficients, 

x'2  +  (2  ;i  -  2)  x'  +  4  ?/'2  +  (8  yfc  —  24)  y'  +  (/i2  +  4  ^-2  -  2  A  -  24  A:  +  21)  =  0. 

If  the  curve  is  symmetrical  with  respect  to  the  new  axes,  there  can  be  no 
terms  of  the  first  degree  in  the  new  equation.     This  is  the  case  if 

27i  -  2  =  0  and  8  A;  -  24  =  0,  or  A  =  1,  k  =  3. 

Hence  the  center  of  symmetry  is  (1,  3)  and  the  transformed   equation 
becomes  x'2  +  4  ?/'2  —  16  =  0  as  before,  on  substituting  h  =  1,  ^  =  3. 


*  In  solving  for  x,  the  terms  not  involving  x  may  be  disregarded,  as  we 
are  interested  only  in  showing  that  the  solution  is  of  the  form 

x  =  h±/{y). 


TRANSFORMATION   OF   COORDINATES         113 


Example  2.  —  Simplify  y^  -{- Sx  + 'iy  -  20  =  0. 

Solution.  —  If  the  equation  is  solved  for  x  no  axis  of  symmetry  is  re- 
vealed, and  so  we  must  use  the  second  method.  Put  z  =  x' -\- h,  and 
y  =  y'-\-k.  Collecting  coefficients,  the 
equation  becomes 

?/'2  +  8x'+  (2;t  +  4)?/'  + 

(A-2  +  8/?. +  4^--20)  =0. 

Evidently  the  terms  in  y'~  and  x' 
cannot  be  eliminated.  But  if  2  A:  +  4  =  0 
and  A;2  +  8  A  +  4  ^  —  20  =  0,  the  terms  in 
y'  and  the  constant  term  vanish.  These 
equations  give  k  =—2,  h  =  3.  The  new 
equation  then  is 

y'2=-Hx', 

which  is  of  the  form  y^  =—2px.  The 
curve  is  a  parabola  which  has  its  vertex 
at  (3,  —  2)  and  its  axis  y  =—2. 


PROBLEMS 

^  1.    Find  the  center  of  symmetry  by  solving  for  x  and  y,  and  move 
the  origin  to  this  point.     Draw  the  curve  and  both  sets  of  axes. 

(a)  9  x2  -  36  X  +  4  2/2  +  16  ^  +  16  =  0  ; 

(b)  16  x2  -  64  a:  -  25  2/2  -  336  =  0  ; 

(c)  4  x2  -  ?/2  +  32  a:  +  4  ?/  +  60  =  0  ; 
/(d)  a;2  ^_  y2  _  18  X  +  6  y  +  54  =  0  ; 

(e)  x^-\-x  +  3y'--9y  +  ^  =  0; 

(/)  6  ic2  -  4  </2  _  12  X  +  16  ?/  +  34  =  0. 


<   2.    Simplify  the  following  equations  by  the  seeond  method.     Draw 
the  curve  and  both  sets  of  axes. 

(a)  4 a:2  +  2/2  +  16  X  -  6  ?/ =  12  ; 

(6)  I6x^  -  y^' +  S2x- 6y  =  0; 

(c)  4x^  -4:X-Sy-3  =  0; 

-id)  y^-^6y-3x-\-Q  =  0; 

(e)  3  a;2  -  18  X  +  4  ?/2  _  8  ?/  -  5  =  0  ; 

(/)  4x2  + 16x +  3?/2- 12?/  + 12  =  0; 

(gr)  !/  =  x3  +  9  x2  +  27  X  +  27  ; 

(h)  y  =  x3  -  4  x2  +  3  X. 


114  ANALYTIC    GEOMETRY 

s/   3.    Find  the  equation  of  each  of  the  following  conies  and  translate 
the  origin  to  the  center  of  symmetry  : 

(a)  Focus  (4,  0),  directrix  x  =  0,  e  =  | ; 
?^b)  focus  (4,  0),  directrix  x  =  0,  e  =  | ; 
^(c)  focus  (1,  2),  directrix  x  =—  4,  e  =  ^  ; 
(d)  focus  (2,  2),  directrix  x  =  —  4,  e  =  2  ; 
~/^(e)  focus  (3,  —  4),  directrix  y  =  0,  e  =  f  ; 
(/)  focus  (3,  —  4),  directrix  y  =  0,  e  =  ^ ; 
Ig)  focus  (—2,  3),  directrix  ?/  =  1,  e  =  i ; 
(h)  focus  (—2,  3),  directrix  y  =  1,  e  =  2. 

Ans.  (a)  25  x2  +  45 1/2  =  576. 

y~  4.    Find  the  equation  of  each  of  the  following  parabolas  and  trans- 
late the  origin  to  the  vertex  : 

(a)  Focus  (4,  0),  directrix  x  =  0; 
/.{b)  focus  (1,  2),  directrix  x  =—  4  ; 

(c)  focus  (3,  —  4),  directrix  y  =  0  ; 

(d)  focus  (—  2,  3),  directrix  y  =  1  ; 

(e)  focus  (0,  0),  vertex  (0,  2); 
(/)  focus  (-  2,  3),  vertex  (3,  3); 
(gr)  focus  (2,  1),  vertex  (1,  1), 

Ans.   (a)  y'''  =  8x. 

5.   Show  that  the  equation  of  the  conic,  with  the  origin  at  the  focus 
and  the  principal  axis  as  the  x-axis,  is  x'^  +  y'^  —  e^(x  +  p)^. 
Transform  this  to  the  standard  forms  as  in  §  87  for 
(a)  the  parabola;  (6)  the  ellipse  ;  (c)  the  hyperbola. 

In  each  of  the  following  locus  problems  simplify  the  equation  by  a 
translation  of  the  axes  : 

>/  6.    Find  the  locus  of  the  middle  points  of  chords  of  the  ellipse 
rrf^(}x2  Jf-  9  y^  =  144  drawn  from  one  end  of  : 

^(a)  the  major  axis  ;  (b)  the  minor  axis. 

7.  Find  the  locus  of  the  centers  of  all  circles  tangent  to  the  circle 
x2  +  ?/2  —  q2  and  the  line  y=b. 

8.  From  one  focus  of  the  ellipse  bh:^  +  a-y-  =  a-b"^^  focal  radii  are 
drawn  and  bisected.  Show  that  the  locus  of  the  points  of  bisection  is 
an  ellipse,  and  find  its  center  and  foci. 

/    "    90.-  Discussion    of    the  Equation    Ax""  +  Cy^  -{-  Dx -\- Ey -\- 

I       F  —  0.     This  represents  the  general  form  of  a  quadratie  in 
X  and  y  with  the  xy  term  lacking.     The  only  restriction  on 


TRANSFORIVIATION   OF  COORDINATES        115 

the  coefficients  is  that  A  and  C  cannot  both  be  zero.  We 
desire  to  find  under  what  conditions  the  locus  is  a  parabola, 
an  ellipse,  or  a  hyperbola. 

Case  1.  When  either  A  or  C  is  zero.  Suppose  ^  =  0. 
Here  the  equation  has  the  form  yvv  '*  C. 

If  D  =^  0,  by  a  proper  translation  we  can  remove ''^  the  y 
term  and  the  constant,  and  obtain  an  equation  of  the  form 
Cy"^  -\-  Dx'  =  0,  of  which  the  locus  is  a  parabola. 

If  i>  =  0,  the  y  term  can  be  removed  by  translation,  and 
we  get  Cy^  -\-  F'  =  0,  of  which  the  locus  is  imaginary  if 
F'  and  C  have  the  same  sign  ;  the  new  a"-axis  if  i^'  =  0 ;  and 
a  pair  of  lines  parallel  to  the  cc-axis  if  F'  and  C  have  unlike 
signs. 

A.  similar  discussion  holds  when  C  =  0. 

Case  2.  Wien  A  and  C  are  of  like  sign.  Removing  the 
first  degree  terms  by  translation,  we  have 

Ax'^  +  Cy'^  =  F'. 

This  is  evidently  an  ellipse  if  F'  has  the  same  sign  as  A 
and  C  ;  a  point  if  i^"'  =  0  ;  and  an  imaginary  locus  if  F'  has 
the  opposite  sign  to  that  of  A  and  C. 

^  Case  3.  Wlien  A  and  C  are  of  urdike  sign.  Removing 
the  first-degree  terms,  we  have 

A.x'^  +  Cy'^  =  F'. 

This  is  evidently  a  hyperbola  unless  F^  =  0,  in  which  case 
the  locus  is  a  pair  of  lines  intersecting  at  the  new  origin. 

*  By  actual  substitution  of  x  =  x'  +  ^  and  tj  =  ij'  +  k,\t  can  be  shown 
that  (o)  translation  does  not  affect  the  coefficients  of  the  highest  powers 
of  the  variables  ;  (b)  that,  if  neither  A  nor  C  is  zero,  the  first-degree 
terras  can  be  removed  ;  (c)  that,  if  one  variable  appears  only  to  the  first 
degree,  the  first  power  of  the  other  variable  and  the  constant  terra  can  be 
removed  ;  (d)  that,  if  one  variable  is  missing,  the  first  power  of  the  other 
can  be  removed. 


116  ANALYTIC    GEOMETRY 

91.  Limiting  Forms.  —  We  have  seen  in  §  §  58,  70,  83  how 

the  various  conies  may  be  formed  by  passing  a  plane  through 
a  cone.  In  the  previous  section  we  have  found  that  in  gen- 
eral the  locus  of  the  equation  Ax^  -\-  Cy"^  -\-  Dx -\-  Ey  +  F  =  0 
is  a  conic,  but  that  for  particular  values  of  the  coefficients 
the  locus  is  imaginary,  or  is  a  point,  a  line,  or  a  pair  of  lines. 
These  degenerate  cases  may  be  regarded  as  limiting  forms 
of  sections  of  a  cone.  The  discussion  of  the  different  cases 
is  as  follows  : 

(a)  If  the  vertex  of  the  cone  is  moved  away  from  the 
base  indefinitely,  the  cone  approaches  the  cylinder  as  a 
limiting  form.  Here  the  elements  are  parallel  and  the  para- 
bolic section  approaches  a  pair  of  straight  lines.  If  the 
cutting  plane  becomes  a  tangent,  the  limit  is  a  double 
straight  line.  If  the  plane  is  moved  farther,  there  is  no  in- 
tersection, or  an  imaginary  one. 

(h)  If  the  plane  forming  a  circular  or  elliptical  section  is 
moved  towards  the  vertex,  the  section  grows  smaller,  until 
at  the  vertex  it  becomes  a  point.  If  it  is  moved  still  far- 
ther, there  is  no  intersection,  or  an  imaginary  one. 

(c)  Any  section  of  a  cone  formed  by  a  plane  passing 
through  an  element  is  a  triangle,  hence,  if  the  plane  forming 
a  hyperbolic  section  is  moved  to  the  vertex,  the  section  be- 
comes two  intersecting  straight  lines. 

92.  General  Statement.  —  The  preceding  analysis  may  be 
summed  up  as  follows  : 

A  quadratic  equation  of  the  form 

Ax"  +  Cy  +  Dx  -\-Ey-bF=0 

always  defines  a  conic  or  one  of  its  limiting  "forms,  the  coeffi- 
cients being  any  real-numbers,  zero  included. 

If  either  A  or  C  is  zero,  the  conic  is  a  parabola  or  in  special 
cases  two  parallel  straight  lines  distinct,  coincident,  or  imag- 
inary. 


TRANSFORMATION   OF  COORDINATES        117 

If  A  and  C  have  the  same  sign,  the  conic  is  an  ellipse,  or  in 
special  cases  a  circle,  point,  or  imaginary  ellipse. 

If  A  and  C  have  unlike  signs,  the  conic  is  a  hyperbola,  or  in 
special  cases  two  intersecting  straight  lines. 

Ji^J^  93.   Generalized  Standard  Equations  of  the  Conies.  —  The 

equations  derived  in  Chapters  IV,  V,  and  VI  we  have  seen 
to  be  of  a  special  character,  'since  they  are  applicable  only 
to  the  case  that  the  vertex  or  center  is  at  the  origin  and  the 
principal  axis  is  the  x-  or  ^/-axis.  We  are  now  in  a  position 
to  generalize  these  equations  and  obtain  standard  forms 
where  the  only  restriction  is  that  the  principal  axis  is  paral- 
lel to  one  of  the  coordinate  axes. 

Consider  an  ellipse  with  the  center  at  (h,  k),  semi-axes  a 
and  b,  and  transverse  axis  y  =  k.  It  is  required  to  find  its 
equation. 

If  the  origin  is  translated  to  (h,  k),  we  know  from  Chap- 
ter V  that  the  transformed  equation  will  be 

/y.'2  ,,'2 

—  4-^  =  1. 
a"       62 

The  problem  then  is  the  reverse  of  that  considered  in  §  89, 
viz.,  the  transformed  equation  is  given  and  the  equation 
with  reference  to  the  old  axes  is  to  be  found.  Hence  the 
substitutions  are  those  of  §  85  reversed,  that  is,  x'  =  x  —  h, 
and  y'  =  y  —  k.     These  give 


'  as  the  required  equation.  \         V ' '-    r        Vi 

In  the  same  way  we  discuss  the  other  cases  and^bta^i 
the  following  equations  :  ^ 

Parabola  — 

'Vertex  (/i,  k),  axis  y  =  k\     (y-ky~  =  2  p(x  -  h)  ;  (30) 

"  "         "      x=:h:     (x-hy  =  2p(y-k);  (30a) 


lla 


ANALYTIC   GEOMETRY 


Ellipse  — - 


Center    (h,  fc),  axis  ?/  =  A; : 


_,.    (x-hf  ,  (y-kf 


+ 


(31) 


((  a 


X 


=  A:     (y^Z^  +  iA^JH^l;       (31a) 


Hyperbola—  Tz/ -  *)^ 

Center    (7^%  axis  ?/ =  A- :    i:^^^-iL^  =  l;         (32) 


^  =  n:     (K^:^^(AIZ31^1.        (32a) 


6» 


These  forms  are  easy  to  remember  if  one  bears  in  mind 
that  they  are  merely  generalizations  of  the  simple  forms 
y'^  =  2px,  x-  =  2py,  etc.  To  reduce  an  equation  to  orie  of 
these  forms,  it  is  necessary  only  to  complete  the  squares. 


Example  1.  —  Write  the  equation  of  the  ellipse  which  has  the  center 
(0,  1),  one  focus  (—4,  1),  and  minor  axis  6. 

Solution. —Here   the   principal  axis  is  y  =  l.    From  the  given  data 
&  =  3  and  c  =  i,  whence  a  =  5.     Therefore  the  equation  is 

25  9 

Example  2.  —  Simplify  the  equation  2x^'  -  Sy'^  -\-8  x  -  G  y  -h  11  =  0. 
Find  the  center,  semi-axes,  vertices,  and  foci,  and  draw  the  curve. 

Solutio7i.  —  Completing  squares, 
2(a:  +  2)2-3(2/  +  l)2=-6 

3  2 

W[    The  center  is  (—  2,  —  1)   and  since   the 
^    term  in  {y  + 1)''^  is  positive,   the  trans- 
verse   axis    is    X—  —  2.     Hence    a=\2, 
b  =  V3,  and  c  =  Vs,    The  vertices  are 


and  the  foci     (-2,  —  1±V5)> 


TRANSFORMATION   OF   COORDINATES         119 


PROBLEMS 

^  1.   Write  the  equations  of  the  following  ellipses  in  the  general  form, 
the  major  axis  being  first  parallel  to  the  x-axis,  and  second  parallel 
to  the  y-Rxis,  unless  the  given  data  fix  its  direction, 
(a)  a  =  5,  ?>  =  3,  center  (2,  3)  ; 
(6)  a  =i,b  =2,  center  (-1,2); 
y  (c)   a  =  f,  6  =  I,  center  (-  2,  3)  ; 
(d)  vertices  (±5,  3),  focus  (3,  3)  ; 
-7Cj(e)  vertices  (—  3,  1),  (7,  1),  focus  (0,  1)  ; 
(/)  foci  (±3,  -  2),  directrix  x  =  —  4 ; 
(gr)  minor  axis  12,  e  =  f^g,  focus  (0,  0)  ; 
s^  (A)  major  axis  6,     e  =  ^,  focus  (1,  1)  ; 

Ans.     id)  5;  +  (^^=l; 
2o  lb 

^^^        169  36 

4 

2.  Write  the  equations  of  the  hyperbolas  satisfying  the  data  of 
Problem  1,  if  possible.  If  not  possible,  explain  why  and  revise  data  to 
make  a  similar  problem  for  the  hyperbola  and  solve. 

Ans.     (d)  not  possible,  for  c<a.     Let  the  vertices  be  (i  3,  3)  and 

the  focus  (5,  3).     Then  the  solution  is  —  -  iMn^):  =  i. 
^  '    ^  9  16 

3.  Write  by  inspection  the  equations  of  the  parabolas  having: 

(a)  focus  (0,  0),  vertex  (2,  0);       (c)  focus  (-2,  3),  vertex  (3,  3). 
(6)  focus  (2,  1),  vertex  (2,  5)  ; 

4.  Simplify  the  following  equations  by  completing  the  squares  and 
draw  their  loci : 

(a)  3x2-2  2/2-6x-8?/-ll  =  0;  (/)  Sy  =  (x  -  1)  (x  +  2) ; 

^b)  x^^4x-by-6  =  0;  (g)  y^  =  (x  -  1)  (x  +  2)  ; 

(c)  4  X-  -  9  y2  +  12  X  +  6  ?/  =  28 ;  (/i)  x^  4-  4  ?/2  +  6  x  +  4  ?/  =  0  ; 

Id)  9?/2-  12?/  -2x-2  =  0;  (i)  4x2  +  ?/2  +  4x  +  6?/ =  0  ; 

(e)  9x2  +  4^2  _36x  +  4y  + 1=0;  (j)  (2x -1)2  +  (3y  -  2)2  =9y2. 

5.  Write  the  solutions  of  Problems  6,  7,  and  8,  page  114,  in  standard 
form. 

6.  The  base  of  a  triangle  is  fixed  in  length  and  position.  Find  the 
locus  of  the  vertex  if  one  base  angle  is  twice  the  other. 

94.  Simplification  of  Equations  by  Rotation.  —  The  process 
of  rotation  of  the  axes  effects  the  removal  of  the  xv  term 


120 


ANALYTIC   GEOMETRY 


from  the  second-degree    equation  if  the  proper  angle  0  is 
chosen.     Let  us  substitute  in  the  general  equation 

Ax'  +Bxy  +  Cf  +  Dx  +  Ey  +  F=  0 
the  rotation  formulas 

X  =  x'  cos  0  —  y'  sin  6, 
y  =  xf  sin  Q  -\-  y'  cos  Q. 
The  coefficient  of  the  x^y'  term  will  be 
-  2  ^  sin  ^  cos  ^  +  5  ( cos^  Q  -  sin^  ^;  +  2  C  sin  (9  cos  d 
or  -A  sin  2^ +  5  cos  2^+  C  sin  2  (9. 

(The  student  should  make  the  substitution  in  full  for  all 
terms  of  the  general  equation.) 
The  term  in  x^y^  will  vanish  if 


.(C-.4)  sin2^  +  5cos2^  =  0,    or     tan  2  0  = 


B 


(33) 


A-C 

Thus  we  have  the  theorem  : 

For  ctny  second  degree  equation  there  is  an  angle  of  value 
less  than  90°  such  that  the  substitutions  for  rotation  through 
this  angle  will  transform  the^  equation  into  one  containing  no 
xy  term. 


For  the  equation  tan  2  ^  = 


B 


gives  values  of  tan  2  6, 


A-C 

positive  or  negative,  from  0  when  5  =  0  to  infinity  when 
A=  C.  Hence  for  all  values  of  A,  B,  and  C,2  0  will  have 
some  value  between  0  and  180°. 

Example.  —  Simplify    41  x-  —  '2^xy-\-  34  if=  25. 

Here  A  =  41,  B  =  -  24,  and  C=  34. 
-24 


Therefore      tan  2  6  = 

This  gives  at  once 

cos  2  ^  = 


7 

-7 
25 


Substituting  in  the   half   angle   for- 
mulas of  trigonometry, 


sin  6 


-^^ 


—  cos  2  6 
2 


4 
5' 


r^ 


TRANSFORMATION   OF   COORDINATES         121 


and  cos^  =  A— ^^ =-• 

2  5 


=V^ 


Q  A  4  Q 

Hence  x  =-x' y'  and  y  =~  x'  -\--  y'- 

5  5  ^5  5 

Substituting  in  the  above  equation  and  reducing,  we  get 

xi'2  +  2  y'-^  =  1, 

an  ellipse  of  semi-axes  1  and 

2 

95.   The  General  Equation  of  the  Second  Degree.  —  We  have 

seen  that  the  equations  of  the  conies  are  all  special  cases  of 
the  general  second-degree  equation 

Ax^  +  Bxy  -\-  Ci/  -\-  Dx  +  Ey  +  F  =  0. 

We  are  now  ready  to  prove  the  converse  theorem ;  namely, 

Eveiy  equation  of  the  second  degree  in  two  variables  defines 
a  conic  or  one  of  the  limiting  forms  of  the  conic. 

It  has  been  shown  in  §  90  that  this  theorem  is  true  for 
every  form  of  the  second-degree  equation  in  which  the  prod- 
uct xy  does  not  appear.  Also  we  have  just  seen  in  the  pre- 
vious section  that  by  a  proper  rotation  of  the  axes  the  xy 
term  can  be  made  to  disappear  from  the  equation.  Thus 
the  proof  of  the  theorem  is  complete. 

^  96.  The  Characteristic.  —  The  quantity  5^  —  4  J.(7  is  called 
the  characteristic  of  the  general  second-degree  equation,  and 
is  denoted  by  A.     We  now  prove  the  theorem  : 

Tlie  characteristic  of  a  general  equation  of  the  second  degree 
is  unaltered  hy  a  rotation  of  the  axes  through  any  angle  6. 

In  the  general  equation  substitute  the  values  given  in  the 
rotation  formulas.     The  first  three  terms  become 


A  cos2  6 
-\-  B  sin  6  cos  0 

+  Csin2^ 


a?'2     -2  A  sin  6  cos  6 

—  B  sin2  e 

-f-  B  cos2  e 

-f  2  C  sin  ^  cos  ^ 


x'y'        +^sin2^ 
—  ^  sin  ^  cos  6 

+   CC0S2^ 


y'' 


122  ANALYTIC   GEOMETRY 

Calling  the  coefficients  of  tlie  new  equation  A',  B',  C\ 
etc.,  we  have : 

A'  =  A  cos2  e  ^-  C  sin2  0 -^  B  sin  6  cos  9, 

C  =  A  sin2  e-\-C  cos2  0-  B  sin  6  cos  0, 

B'  =  2  C  sine  cose  -2  A  sin  e  cos  9  -j-  B  cos^  $-  B  sin^  ^ 
The  characteristic  of  the  new  equation,   B'^  —  4:A'C',   be- 
comes on  multiplying  and  collecting  the  terms  : 
B^  cos^  e-^-B"  sin^  9  -  S  AC  sin^  9  eos^  ^  -  4  .4C  sin^  9  - 

4  ^C  cos-*  9  +  2B-  sin-  ^  cos^  9 

=  (B2  -  4  ^(7)  cos^  ^  +  (52  -  4  AC)  sm'9  +  2(52-4.16') 

sin2  9  cos2  ^ 

=  (52  -  4  ^0)(cos4  (9  4-2  sin^  ^  cos^  ^  +  sin^  9) 

=  (52  -  4  ^C)(cos2  9  +  sin2  ^)2 

=  52_4^(7 

It  is  easy  to  see  that  A  is  also  unchanged  by  a  transla- 
tion of  the  axes.  For  this  reason  it  is  called  an  invariant 
of  the  equation. 

Exercise.  —  Prove  that  ^  +  C  is  not  changed  by  the  substitutions 
for  rotation  or  translation  of  the  axes. 

97.  Test  for  distinguishing  the  Conies.  —  If  in  the  previous 
section,  the  angle  9  were  chosen  so  that  the  xy  term  van- 
ished, the  new  equation  would  be 

A'x'""  +  Cy'"-  +  Dx'  -\-Ei/  -h  F=0 
and  /^  =  B"-  4.  AC  =  B''-  -  4:  A'C  =  -  4.A'C', 

since  B'  =  0. 

From  this  relation  and  §  92  we  see  that 

(/"A  <  0,  A'  and  C  are  of  like  sign,  and  the  conic  is  an 
ellipse* 

if  A=  0,  A'  or  C  is  zero,  and  the  conic  is  a  parabola,* 

if  /\  >  0,  A'  and  C  are  of  unlike  sign,  and  the  conic  is  a 
hyperbola* 

*  Or  one  of  the  limiting  forms  discussed  in  §§  90,  01. 


TRANSFORMATION   OF   COORDINATES         123 

98.    Suggestions  for  simplifying  the  Equation  of  a  Conic.  — 

If  the  conic  is  an  ellipse  or  hyperbola  (that  is,  if  A  ^  0) 
determine  the  coordinates  (h,  k)  of  the  center  of  symmetry 
and    remove   the  tirst-degree    terms  by  translation.     Then 
rotate  the  axes  as  in  the  example  of  §  94. 

If  the  conic  is  a  parabola  (A  =0),  the  substitutions  for 
rotation  should  be  made  first,  and  then  equations  of  condi- 
tion involving  h  and  k  can  be  formed. 

A  convenient  check  on  the  accuracy  of  the  rotation  sub- 
stitutions is  the  fact  that  A  -\-C  is  imchanged  until  the  new 
equation  is  simplified.     (See  Exercise,  §  96.) 

PROBLEMS 

Simphfy  the  following  equations.  In  each  case  draw  the  conic  and 
the  three  sets  of  coordinate  axes. 

1.  73x2  +  72x?/  +  52?/2-218x-  116  y  -f  07  =  0. 

i  Ans.     4x"2  +  y"2  =  4.    . 

2.  -^2  _  24  a-?/ +  11 1/2 +  72  a:- 116?/+ 204  =0. 

Ans.     a:"2  -  4  ?/"2  =  _  4. 

3.  18  x2  -  48  a;?/  +  32  y^  _  120  x  +  35  ?/  +  200  =  0. 

/Ty  .^  -  /,'  ^ -  ■  Alls.     2y"'^  =  Zx". 

4.  6  x2  +  13  x?/  +  6  2/2  -  7  X  -  8  ?/  +  2  =  0.     Ans.     Two  Hnes. 

5.  2  x2  +  4  xy  +  4  ?/2  +  2  X  +  3  =  0.  Ans.     Imaginary. 

6.  41x2 +  24x?/ +  34?/2_  90x+ 120I/  +  225  =  0.     Ans.   Point. 

7.  16  x2  +  24  x?/  +  9  ?/2  -  60  X  -  45  ?/  +  50  =  0.     Ans.  4  x"2  =  1. 

8.  66  x2  +  24  xy  +  59  ?/2  +  396  x  +  72  ?/  +  444  =  0. 

Ans.     3  x"2  +  2  ?/"2  =  6. 

9.  x2  +  24  x?/  -  6  ?/2  +  4  X  +  48  ?/  +  34  =  0. 

Ans.   3  2/"2  -  2  x"2  =  6. 

10.    16  x2  +  24  x?/  +  0  2/2  _  60  X  +  80  ?/  +  200  =  0. 

Ans.     x"2  +  4?/'^  =  0. 

,.•11.    7x2-48x?/-7?/2=0.         ^^ Ans.     x"'^  -  y"-^  =  0. 

12.  2  x2  +  6x?/  +  10  ?/2  -  2  X  -  6  ?/  +  19  =  0.   Ans.     Imaginary. 

13.  9a;2_24x2/ +  16^2_3^_1_  4y  _6  =0.       Ans.     Two  lines. 


124  ANALYTIC    GEOMETRY 

14.  13x-^  +  10 a;?/  +  13  2/2  -  42 X  +  6 ?/  -  27  =  0. 

Ans.     9x"-^  +  4:y"-^  =  S6. 

15.  13x2  + 10x2/ 4-13  y2  +  16  X- 16?/  + 16  =  0.         Ans.     Point. 

16.  x2  +  4x?/ +  ?/2- 2x- 10?/ -  11  =  0.     Ans.    S x"-^  -  y"-^  =9 . 

17.  3  x2  +  8  x?/  -  3  ?/2  _  10  X  -  30  ?/  +  20  =  0. 

Ans.     ?/"2  _  .7.//2  _  4 

18.  6  x2  —  4  X2/  +  9  ?/2  —  40  X  +  30  ?/  +  55  =  0. 

Ans.     x"2  +  2  ?/"2  =  2. 

^    19.    3x2-2V3x?/  +  ?/2_  (8+ 12V3)x+(12-8V3)  ?/- 12  =  0. 

^ns.    ?/"2  =  4x. 

20.  The  ends  of  the  base  of  a  triangle  are  (0,  0  )  and  (4,  0),     Fina 

the  locus  of  the  vertex  if  the  sum  of  the  slopes  of  the  sides  is  {a)  j\, 

2 
(6)1,  (c)  — ~"-      Simplify  the   equation  found  and  draw  the  locus. 

21.  Solve  the  previous  problem  when  the  given  points  are  (0,  0) 
and  (a,  0)  and  the  sum  of  the  slopes  is  any  constant  k.  Then  show 
that  the  locus  is  a  hyperbola  passing  through  the   given  points  and 

2 

having  its  axes  of  symmetry  inclined  at  an  angle  I  arc  tan 

22.  The  difference  of  the  base  angles  of  the  triangle  whose  base 
joins  (0,  0)  and  (4,  0)  is  45°.     Find  the  locus  of  the  vertex. 

Ans.    x"2  -y''-^  =  2V2  or  x"2  -  y"-^=  -  2  V2. 

99.  The  Conic  through  Five  Points.  —  The  general  equation 
of  the  second  degree  involves  six  arbitrary  constants.  As  in 
the  case  of  the  circle,  however,  one  of  these  can  be  divided 
out ;  hence,  if  we  can  express  five  of  the  coefficients  in  terras 
of  the  sixth,  the  conic  is  completely  determined.  Geomet- 
rically, this  means  that  the  conic  is  completely  determined 
by  five  points,  or  by  five  other  geometric  conditions. 

To  find  the  equation  of  a  conic  so  defined  we  proceed  to 
form  equations  between  the  coefficients  of  the  general  equa- 
tion and  solve  them.  For  example,  suppose  the  conic  is  to 
pass  through  the  points  (4,  2),  (2,  4),  (—3,  1),  (1,  —  3),  and 
(0,  0).     Substituting  these  in  the  general  equation, 

Ax"  4-  Bxy  +  CY  +-  Dx  -\- Ey  +  F=  0, 


TRANSFORMATION   OF   COORDINATES        125 

wehave:  16^  +  8^+    4.  C -{- 4.  D  +  2  E  +  F  =  0, 

4.A  +  SB  +  16C+2D  +  AE  +  F=0, 

9A-3B+       C-SD-j-    E  +  F  =  0, 

A-SB+    90+     D-3E-\-F=0, 

F=0. 

Solving  these,  we  find  ^  =  C=  ^,  D  =  E  =  ~  ^.      This 

oO  o 

makes  the  general  equation,  on  dividing  out  the  B  and  clear- 
ing of  fractions : 

x^  -\-  m  xy  +  y-  —  70  X  -  IQ  y  =  0. 

The  value  of  A  shows  that  the  conic  is  a  hyperbola.     Its 
center  is  found  by  the  method  of  §  89  to  be  (ff ,  ff)-     The 
inclination  of  its  axes  is  45°,  since  tan  2  ^  =  oo  .     This  in- 
formation, together  with  the  live  points  given,  is  sufficient ' 
for  drawing  the  curve.    The 
graph  follows. 

A  shorter  method  of  solv- 
ing the  above  problem  is 
given  by  the  theorem  of 
§41.  Call  the  first  four 
points  P,  Q,  E,  and  S.  The 
equations  of  the  lines  PQ 
and  BS  are  x  -\-  y  —  6  =  0 
and  a;  +  2/  +  2  =  0.  ^lul- 
tiplying  these  together,  the 
equation  of  the  pair  of  lines 

is 

x-^  -\-  2  xy  +  y^  —  4:X  —  4: y  -  12  =  0, 

by  §  42.     Similarly  the  equation  of  the  pair  of  lines  RQ  and 
iSP  is 

15  0.-2  -  34  xy  4-  15  /-  +  2Sx-\-  2Sy-  196  =  0, 


126  ANALYTIC    GEOMETRY 

Now  the  intersections  of  these  two  equations  are  the  points 
P,  Q,  R,  and  S.     Hence  by  §  41 

15  x"  -34:xy  -{-  15  if -\- 2S  x  -\- 28  ij  -  196 

+  k(x''-i-2o^!j  +  7f-4.x-4:y-12)  =  0  (a) 

is  the  equation  of  a  system  of  curves  passing  through  these 
four  points.  As  this  is  of  the  second  degree  for  all  values 
of  k,  they  are  all  conies.  To  find  the  conic  of  this  system 
passing  through  the  fifth  point  (0,  0),  substitute  its  coordi- 
nates for  the  variables.     This  gives 

-  196  -12k  =  0, 

whence  k  =  —  ^^-.  Putting  this  value  for  k  in  (a)  and  col- 
lecting terms,  we  have  the  same  result  as  before. 

PROBLEMS 

1.  Find  the  equation  of  the  conic  passing  through  the  following 
points  : 

(a)  (3,  2),  (1,  2),  (3,  0),  (2,  |),  (Y,  1)  ; 

(6)  (1,  6),  (-  3,  -  2),  (-  5,  0),  (3,  4),  (0,  10)  ; 

(c)  (3,  1),  (3,  5),  (6,  2),  (6,  10),  (11,  5)  ; 

(cO  (1,  1),  (-  1,  2),  (0,  _  2),  (-  2,  -  1),  (3,  -3). 

2.  Find  the  equation  of  the  parabola  passing  through  the  following 

points : 

(a)  (0,0),  (4,0),  (0,3),  (-8,8); 

(6)  (0,  1),  (4,  7),  (4,  1),  (12,  7)  ; 

(c)  (1,  1),  (4,0),  (0,4),  (9,  1); 

Hint.  —  One  of  the  conditions  is  B^  —  ^AC  =  0. 

3.  Find  the  equation  of  a  conic  with  axes  of  symmetry  parallel  to 
the  coordinate  axes  and  passing  through  : 

(a)   (-4,5),  (6,0),  (-4,  -5),  (4,3); 
(&)   (7,1),  (6,  -2),  (7,  -o),  (-2,  -2); 
(c)   (-3,  -4),  (-1,  -3),  (1,0),  (.3,5). 

4.  Find  the  equation  of  a  conic  symmetrical  with  resi^ect  to  the 
origin  and  passing  through  : 

(a)  (2,4),  (3,  1),  (-2,0); 
(ft)  (1,2),  (-1,  1),  (3,2); 
(c)   (0,0),  (0,4),  (4,0). 


CHAPTER    VIII 
POLAR  COORDINATES 

100.  Definition.  —  Some  topics  in  analytic  geometry  can 
be  better  investigated  by  the  use  of  polar  coordinates  than 
by  rectangular  coordinates.  In  the  polar  system  the  posi- 
tion of  a  jjoint  is  fixed  by  measuring  a  distance  and  a  direc- 
tion instead  of  by  the  measures  of  two  distances.  This  is 
essentially  the  same  s^^stem  as  that  of  bearing  and  distance 
used  in  surveying,  contrasted  with  that  of  latitude  and 
longitude  used  in  geography. 

Choose  a  fixed  point  0  as  the  origin,  called  the  pole,  and 
a  fixed  line  OA  through  it,  called  the  polar  axis.     Then  any 


"  \>^ 


point  P  is  determined  if  we  know  its  distance  from  0  and 
the  angle  that  OP  makes  with  OA.  The  measures  of  the 
distance  OP  and  the  angle  AOP  are  called  the  polar 
coordinates  of  P  and  are  designated  b}^  p  and  6.  The 
distance  p  is  called  the  radius  vector  of  P,  and  0  is  called  the 
vectorial  angle. 

Polar  coordinates  do  not  obey  the  conventions  of  the 
rectangular  system  as  to  their  direction  and  magnitude.  As 
in  trigonometry  the  radius  vector  may  be  rotated  indefi- 
nitely in  a  counter-clockwise  or  clockwise  direction,  mak- 
ing 6  take  on  any  positive  or  negative  value.  Distances 
measured  on  the  terminal  line  of  6  away  from  the  pole  are 

12.1 


128 


ANALYTIC    GEOMETRY 


positive;  those  measured  in  the  opposite  direction,  on  the 
terminal  produced,  are  negative.  (See  the  right-hand  dia- 
gram on  page  127.)  Hence  every  pair  of  real  numbers  (p,  6) 
determines  one  point  "vrhich  may  be  located  according  to  the 
following  rule. 

EuLE  FOR  Plotting.  —  Taking  the  ptolar  axis  as  an  initial 
line,  lay  off  the  vectorial  angle  6,  counter-clockwise  if  positive, 
clockwise  if  negative.  Then  measure  off  the  radius  vector  p, 
on  the  terminal  of  6  if  positive,  on  the  terminal  of  0  produced 
through  the  pole  if  negative. 

Since  6  and  ^  -|-  2  tt  have  the  same  terminal  line,  a  point 
may  he  represented  by  an  indefinite  number  of  pairs  of 

coordinates.     Thus  in  the  ad- 
^^"'^  joining  figure  ws  may  take  for 

the   coordinates  of   P,   p  =  8, 


±  2  mr.     Since 


$  =  — -    or  — 
6  6 

the  terminal  line  of  0  pro- 
duced is  the  terminal  of  ^  ±  tt, 
a  second  set  of  coordinates  is 


P  = 


'=1 


TT 


or    -  ±  2  nir.     Ordinarily  we  keep  6  within 
6 


the  limits  ±  tt,  or  0  and  2  tt. 

101.   Relations  between  Rectangular  and  Polar  Coordinates. 

—  Take  the  pole  at  the  origin  of  rectangular  coordinates 
and  the  polar  axis  as  the  positive  half  of  the  ic-axis.     Erom 


POLAR  COORDINATES 


129 


the  figures  and  the  definitions  of  the  trigonometric  func- 
tions it  is  evident  that  for  P(x,  y)  =  P(p,  0)  in  any  quadrant 
the  following  formulas  are  true : 

X  =  p  cos  6,  y  =  p  sin  9,  (34) 

^^  p2=:x'^  +  y\  e  =  arctan^.  (35) 

These  equations  enable  us  to  transform  rectangular  equa- 
tions and  coordinates  into  polar  forms,  and  conversely. 

NoTE/  —  In  transforming  the  rectangular  coordinates  of  a  point  into 
polar  coordinates,  care  should  be  taken  to  group  together  the  correspond- 
ing values  of  p  and  d. 

102.  Polar  Curves.  —  The  definitions  of  equation  and  locus 
in  polar  are  the  same  as  those  in  rectangular  coordinates, 
if  (p,  6)  is  substituted  for  (x,  ?/)  (§  14).  The  equation  in  polar 
coordinates  is  derived  as  in  the  rectangular  system  (§  16). 
In  a  few  cases  the  polar  equation  may  be  obtained  best  by 
deriving  it  in  the  rectangular  form  and  then  substituting 
for  X  and  y  their  values  in  terms  of  p  and  6  and  vice  versa. 

Plotting  in  polar  coordinates  resembles  that  in  rectangular 
coordinates.  The  equation  should  usually  be  solved  for  p 
and  a  table  of  values  formed,  taking  values  of  6  at  intervals 
of  30°  (or  sometimes  15°).  When  the  curve  has  symmetry, 
it  is  usually  unnecessary  to  carry  the  table  through  more 
than  two  quadrants. 

Example.  — Plot  the  ellipse  2  p  —  p  cos  0  =  6. 

Solving,    p  =  - — ^ — -. 
2  —  cos  6 

Table   of  Values 


e° 

p 

e° 

p 

0 

6.0 

210 

2.1 

30 

5.3 

240 

2.4 

60 

4.0 

270 

3.0 

90 

3.0 

300 

4.0 

120 

2.4 

330 

5.3 

150 

2.1 

360 

6.0 

180 

2.0 

300 


130  "I.  ANALYTIC    GEOMETRY 

,'S6(!^^^  PROBLEMS 

V  1.    Plot  the  points:  (a)   (0,  ±  30°)  ;   (b)   (±10,420^);  (c)  ^8,  M 

(8,  ~Y     What  symmetry  has  each  pair  of  points  ? 

2.  Plot  ( 8,  — ^  j .     Plot  points  symmetrical  to   this  with   respect 
to  the  pole,  polar  axis,  and  the  90°  axis,  and  find  their  coordinates. 

3.  As  in  Problem  2  find  the  points  symmetrical  to  : 

(«)(ie,i^');W    (-6,-2-');(e)(-10,-?^). 

4.  Fix  P(p,  6)  in  any  quadrant  and  find  its  symmetry  to : 

(a)  P  (/),  -  d)  ■    {b)  P(-  p,  6)  ;    (c)  P(p,  TT  -  6). 

5.  Find  the  rectangular  coordinates  of  the  following  points  : 

(.)      (10,^);  (.)(4,^f);  (c)(6,|); 

(d)   (4,.);  (e)    (-2,^);  (/)    (-8,^). 

Ans.     (a)  (8\/8,  8). 

6.  Find  two  pairs  of  polar  coordinates  for  each  of  the  following 
points  and  plot  the  point  in  each  case : 

(a)   (-1,  Vr);      _     (&)    (2,  -2);  (c)    (5,  -12); 

(d)   (2,  -2V3);  (e)   (-V2,  -  V2). 

ylns.     («)   (2,  120^),  (-2,  300"). 

•j^  7.   Plot  each  of  the  following  equations  and  identify  each  curve  by 
.  W\  ,   transforming  its  equation  to  rectangular  coordinates  : 
\\/ /  (a)  p'=  4  ;  '  (d)  pcos0=^  -2;  (g)  p  =  2  a  cos  9  ; 

(b)  ^  =  arc  tan  I  ;  (e)  p  =  ±Q]  {h)  p  sin  0  =  2  a. 

(c)  p  =2  cos  0  ;  (/)     tan  0  =  §. 

<  8.    Draw  each  of  the  following  curves  and  transform  its  equation 
into  polar  coordinates : 

(a)  y^ox;  (g)  x^  +  ?/  +  4  2/  =  0  ; 

(h)  Sx  +  2y  =  0;  (h)  x^  -  y'^  =  IQ  ; 

(c)  x-a  =  0;  (i)  y^  -  x^  =  16; 

\W  V     <id)  y  =-^',  U)  ^'  -  2/'  =  a' ; 

\^    /         ..  (e)  x2  +  y2  ^  a2  ;  ,^  (k)  2xy  =  a^ ; 

(/)  r.2  +  ?/2  -  4  X  =  0  ;  (Z)  2  .r?/  =  -  a^. 

J.>?,£.     (a)  tan  ^  =  3  ;     (c)  p  cos  ^  =  a  ; 
■(/)  p  =  4cos0;    (h)  p^  cos  2  ^  =  16  ; 
{k)  /)2  sin  2  ^  =  a^. 


POLAR   COORDINATES 


131 


p{p,e) 


103.    The   Equation    of   the    Straight  Line.  —  The   general 
equation  of  the  straight  line  in  polar  coordinates  is  not  as 
convenient  as  the  equation  in  rectangular  coordinates  and 
will    not    be    discussed.      The 
special  cases  where  the  line  is 
parallel  or  perpendicular  to  the 
polar  axis  or  passes  through  the 
pole  lead  to  very  simple  equa- 
tions. 

In  the  figure  let  the  line  /  be 
perpendicular  to  the  polar  axis 
OA  and  have  the  polar  intercept 
a.     For  any  point  F{p,  6)  on  the 

line  it  is  evident  that  cos  6  =,  whence  the  equation  of 
the  line  is  ^ 


p  cos  6  =  a. 


(36) 


Similarly  the  equation  of  a  line  parallel  to  the  polar  axis, 

of  90°  intercept  a,  is 

p  sin  6  =  a.  (36a) 

For   a   line    passing    through   the    pole    the    equation  is 

evidently 

e  =  c.  (37) 

Exercise  1.  By  transforming  the  normal  form  of  the  straight-line 
equation  into  polar  coordmates  show  that  the  equation  of  any  line  in 
polar  coordinates  is 

p  cos  (6  —  u)  =  p. 

104.  The  Equation  of  the  Circle.  —  As  in  the  case  of  the 
straight  line  the  general  form  of  the  circle  equation  is  not 
often  used.     Several  special  forms,  which  are  common,  are : 

Circle  icith  center  at  pole,  radius  r :       p  =  '' ;  (38) 

Circle  imtli  center  at  (/•,  0),  radius  r :     p  =  2r cos  9  ;        (39) 

Circle  ivith  center  at 


TT 

''  o  ;? 


2 


radius  r:  p  =  2  r  sin  6  ;     (39a) 


n 


132 


ANALYTIC    GEOMETRY 


Circle  through  pole,  with  polar  intercept  a  and  90°  inter 
cept  b : 


p  =  a  cos  0  +  &  sin  8. 


(40) 


p(p.e) 


The  first  equation  is  ob- 
vious. The  second  may  be 
readily  derived  frora  the  ad- 
joining figure,  and  the  third 
from  a  similar  figure. 

The  simplest  way  of  getting 
the  fourth  equation  is  to  ob- 
serve that  its  rectangular  equa- 
tion is  X-  -\-  y~  —  ax  —  by  =  0. 
Transforming  this  into  polar 
coordinates,  we  have  the  de- 
sired result. 

Exercise  2.     Derive  equation  (40)  directly  from  a  figure. 

Hint.  —  Show  that  the  radius  vector  is  in  length  the  sum  of  the  projec- 
tions upon  it  of  the  x-  aud  y-  intercepts  of  the  circle. 


105.  Discussion  of  Polar  Curves.  —  As  in  the  case  of  rec- 
tangular equations,  the  study  of  polar  curves  is  facilitated 
by  discussion  of  the  equation  in  conjunction  with  plotting. 
The  topics  discussed  are  of  much  the  same  character. 

(a)  Intercepts.  —  The  intercepts  on  the  polar  and  90° 
axes  are  the  values  of  p  for  ^  =  0°,  90°,  180°,  and  270°. 

(6)  Symmetry.  —  It  is  easy  to  establish  that  in  accord- 
ance with  the  definitions  of  §  22,  (p,  6)  is  symmetrical  to 
(—  p,  6)  with  respect  to  the  pole ;  to  (p,  —  9)  with  respect 
to  the  polar  axis  ;  and  to  (p,  tt  —  6)  with  respect  to  the  90° 
axis.     Hence  we  have  the  following  tests  for  symmetry : 

If  the  substitution  of 

—  p  for  p  (  does  not  change  the  form  of  the  1  the  p)ole; 

—  6  for  6  equation,  there  is  symmetry  icith  \  the  polar  axis; 
IT  —  6  for  6  [  respect  to  J  the  90°  axis. 


POLAR  COORDINATES  133 

Exercise  3.  Prove  that  (p,  ^)"is  symmetrical  to  : 

(^(i)  ( —  p,  ^)  with  respect  to  the  pole ; 

(5)  (p^  _  ^)  with  respect  to  the  polar  axis  ; 

(c)  (p,  TT  —  ^)  with  respect  to  the  90'^  axis ; 

(d)  (p,  TT  +  ^)  with  respect  to  the  pole  *  ; 

(e)  (_  p,  TT  _  ^)  with  respect  to  the  polar  axis*  ; 
(j^  (_  p,_  ^)  with  respect  to  the  90°  axis.* 

(c)  Limiting  Values  of  O.  —  li  the  solution  for  p  gives 
p  =  a±  Vj\0j,  the  equation  /((9)  =  0  usually  determines  limit- 
ing values  of  6,  since  values  of  0  making  f(0)  <  0  make  p 
imaginary  and  must  be  excluded. 

(d)  Extent  of  the  Curve.  —  If  any  values  of  0  make  p  in- 
finite, these  values  determine  the  direction  in  which  the 
curve  extends  to  infinity.  If  p  is  never  infinite,  the  values 
of  6  which  make  p  take  on  its  greatest  numerical  value 
should  be  found.  Likewise  the  values  of  0  for  which  p  =  0, 
or  for  which  p  takes  on  its  least  numerical  value,  should  be 
found. 

Example    1.  —  Discuss  and  plot  the   locus   of  the   equation  p  = 
2 


e 

0° 

90° 

180° 

270° 

p 

1 

2 

CO 

2 

1  +  cos  6' 

(a)  Intercepts 

(&)  Symmetry  :  With  respect  to  the  polar  axis,  since  cos  (—  0)  = 
cos  e.     Hence  the  table  of  values  is  unnecessary  for  6  >  180°. 

(c)  Limiting  values  of  6  :     None,  since  no  radical  is  involved. 

(d)  Extent :  p  is  infinite  when  1  +  cos  ^  =  0,  i.e.  when  cos  0  =  -  1, 
or  ^  =  TT. 

p  is  evidently  never  0.  Its  least  value  occurs  when  1  -h  cos  ^  is 
greatest,  i.e.  when  cos  ^  =  1,  or  ^  =  0.     Here  p  =  1. 

*  Exercise  3,  d,  e,  and  /,  gives  three  alternative  tests  for  symmetry. 
Their  existence  is  due  to  the  double  representation  of  a  point  in  polar 
coordinates. 


134 


ANALYTIC    GEOMETRY 

Table  of  Values 


e 

(radians) 

0 

(degrees) 

1  +  cos  fl 

p 

±0 

0 

2.00 

1.0 

~  6 

±     30 

1.87 

1.1 

^3 

±    60 

1.50 

1.3 

TT 

"^2 

±    90 

1.00 

2.0 

'    3 

±  120 

0.50 

4.0 

^   6 

±  150 

0.13 

15.0 

—    TT 

±  180 

0.00 

X 

Ti'ansformation  to  rectangular  coijrdinate.s  shows  that  this  curve  is 
a  parabola. 


Example  2.  —  p-  =  a-  cos  2  6,  or  p  =  ±  aVcos  2  d. 

(a)  The  intercepts  on  the  polar  axis  are  ±  a  ;  on  the  90^  axis  they 
are  imaginary. 

(6)  The  three  tests  for  .symmetry  are  satisfied,  for  cos2(-P) 
=  cos2  ^,  etc.     Hence  vakies  of  d  up  to  90"  only  are  needed. 

(c)  p  is  imaginary'  when  cos  2  ^  <  0,  i.e.  when 

90^  <  2  ^  <  270=, 
or  45=  <     ^  <  135=. 

(d)  The  greatest  numerical  vahie  of  p  occurs  when  cos2^=l,  or 
e  =  0  or  180=.     This  value  is  a. 

p  =  0  when  cos  2  ^  =  0,  or  when  2  ^  =  90=,  270=,  etc.,  or  6  =  45=, 
135=,  225=,  and  315=. 

Table  of  Values,  (a  —  10) 


e 

•2  0 

cos  2  e 

p 

Vcos  -.i  e 

0 

0 

1.00 

±1.00 

±10 

15= 

30 

.87 

±    .93 

±  9.3 

30= 

60 

.50 

±    .71 

±7.1 

45^ 

90 

.00 

.00 

0.0 

POLAR   COORDINATES 


135 


Example  3.  —  p  =  1  +  2  sin  ^. 


(a) 


0° 


1 


90° 


180^ 


1 


270° 


-  1 


(h)  Since  sin  (tt-^)  =  sin  d,  the  curve  is  symmetrical  with  respect  to 
the  90°  axis. 

(c)  No  values  of  6  are  excluded. 

{d)  p  is  never  infinite;  its  greatest  value  occurs  when  sin  ^  =  1,  or 
e  =  90°. 

p  ^0  when  1  +  2  sin  ^  =  0,  or  sin  ^  =  -  ^.  Hence  the  curve  passes 
through  the  pole  when  ^  =  210^  or  330°. 


Table  of  Values 


9 

sin  9 

p 

0 

.00 

1.00 

30 

.50 

2.00 

60 

.87 

2.73 

90 

1.00 

3.00 

120 

.87 

2.73 

150 

.50 

2.00 

180 

.00 

1.00 

0 

sin  9 

p 

210 

-.50 

.00 

240 

-  .87 

-  .73 

270 

-  1.00 

-  1.00 

300 

-.87 

-  .73 

330 

-.50 

.00 

360 

.00 

1.00 

PROBLEMS 

1. 

Identify  and  draw  the  locus  of  each  of  the  following : 

(a) 

p  cos  e  —  4  =  0  ; 

(g)  P^  +  3p-4  =  0; 

(&) 

p  cos  ^  +  4  =  0  ; 

(h)  sec2d  =  2; 

(c) 

P-4; 

(0  ^  =  ^; 

id) 

tan  ^  =  V3  ; 

D 

(e) 

p  -  8  sin  ^  =  0  ; 

(J)   p  =  -Q  sine; 

(/) 

/)  +  7  cos  ^  =  0  ; 

(k)  p  =  3  cos^; 

(?)   p-  -  2  p  (cos  0  —  sin  ^)  —  7  = 

2.  Transform  into  polar  coordinates  : 

(a)  Ax  +  By  +  C  =  0; 

(&)  x2  +  y^  -\-  Dx  +  Ey  -\-  F=  0. 

Discuss  and  plot  the  locus  of  each  of  the  following  : 

X  3.    (a)  p  =  a  (1  —  cos  d);      (c)  p  =  a  (1  —  sin  6) ; 

u  lb)  p  =  all  +  cos  d);      (d)  p  =  a  (1 -i- sind).     (Cardioids.) 


136 


ANALYTIC    GEOMETRY 


4.    (a) 


p  =  a  -{-  b  cos  6  ; 
p  =  a  —  b  cos  6  ; 


5.'\a)  p  =  a  tan-  ^  sec  ^  ; 


(c)  p  =  a  +  ti  sin^  ; 
(rZ)  p  —  a  —  b  sin  ^,    (Lima^ons  :  three 
forms  according  as  a  <,  =,  or  >  b.) 

(?))  p  =  a  cot-^  csc^. 

(Semi-cubical  parabolas.) 

6.    (a)  p  =  asece  ±b;  (b)  p  =  acscd  ±  b.     (Conchoids  of 

Nicomedes  ;  three  forms  according  as  a  <,  =,  or  >  6.) 

p  =  2a  tan^sin^;       (?>)  p  =  2  a  cot  ^  cos  ^.     (Cissoids.) 


p  =  a  sin  2  6 
p  =  a  sin  3  6 
p  =  a  sin  5  ^ 


(6)  p  =  «  cos  2  ^.     (Four-leafed  roses.) 
(5)  p  =  a  cos  3  ^.     (Three-leafed  roses.) 
(?))  p  =  a  cos  5  0.     (Five-leafed  roses.) 

a2  sin  2  ^.     (Lemniscate.) 

ad.     (Spiral  of  Archimedes.) 

6«^.     (Equiangular  spiral.) 

=  a.     (Lituus.) 


106.   Rotation  of  Axes.  —  Rotation  of  the  polar  axis  does 
not  affect  the  vahie  of  p.     The  figure  shows  that  if  the  co- 
ordinates of  P  with  refer- 


P(P,0) 

{Pid') 


ence  to  OA  are  (p,  6),  and 
to  0^1'  are  (p,  0'),  while 
angle  ^4' 0-4  is  «,  we  have 
as  a  formula  of  rotation 

e  =  6'  +  a.        (41) 

By  rotating  the  polar  axis 
the    standard  equations  of 
curves  may  be  Avritten  in  a 
variety  of  forms. 
Translation  of  the  pole  is  seldom  necessary  and  is  best 
performed  by  turning  the  equation  into  rectangular  coordi- 
nates. 

,  ♦  Express  e  in  radians, 


POLAR   COORDINATES  137 

107.  Symmetrical  Transformations.  —  Equations  may  also 
be  transformed  by  the  substitutions  for  symmetry.  As  in 
the  case  of  rectangular  coordinates  it  can  be  shown  that  the 
effect  of  such  a  substitution  is  to  transform  the  locus  into  a 
curve  symmetrical  to  the  given  locus  with  respect  to  the 
center  or  axis  involved.     Thus  we  have  the  following  : 


If  in  an  equation  substitution  is  made  of 


—  p  for  p 

-  6  for  6 
TT-e  for  0 


the  locus  of  the  new  equation  " 

is  symmetrical  to  that  of  the 

old  with  respect  to  the 

PROBLEMS 


pole ; 

polar  axis ; 
90°  axis. 


1.  In   the   following  find  the   new  equation  on  rotating  the  axes 
through  the  indicated  angle,  and  draw  curve  and  both  axes  : 

(a)  p  =  a  tan-^  d  sec  6,  u  =  90= ;  (c)  p  =  a  sin  3  6,  a  =  30°  ;     • 

C&)  p  =  asece  ±b,  a  =  90' ;  {d)  p  =  acosd,  a  =  90°  ; 

(e)  p  =  a  -\-b  cose,  a=:  90^. 

Ans.     (rt)  p  —  —  acot:^  dcscd. 

2.  Show  that  the  first  of  each  of  the  following  pairs  of  equations 
may  be  transformed  into  the  second  by  a  proper  rotation  of  the  polar 

axis; 

(a)  p  =  a  {I  +  cosd),  p  =  a  (1  -  sin  ^) ; 

(6)  p  =asin2^,  p  =acos2^; 

(c)  p  =  a  sin  S  d,  p  =  a  cos  3  6  ; 

(rf)  p-  =  a'^  sin  2  6,  p^  =  a-  cos  2  d. 

3.  Find  the  new  equation  when  the  substitution  for  symmetry  with 
respect  to  the  90°  axis  is  made,  and  draw  both  curves  on  the  same  axes  : 

(a)  p  =  a{l  -  cos  d) ;       (c)  p  =  a  tan^  ^  sec  ^ ;        (e)  p  =  a  cos  3  ^  ; 
(6)  p  =  a  +  6  cos  ^  ;        (rf)  p  =  2  a  tan  ^  sec  0  ;     (f)p  =  a  cos  6. 

Ans.     (a)  p  =  «(1  +  cos^). 

4.  To  each  equation  apply  the  substitution  for  symmetry  with  re- 
spect to  the  polar  axis  and  draw  both  curves  on  the  same  axes. 

(a)  p  =  a(l  -  sin  ^)  ;  (c)  p  =  a  sin  3  ^  ; 

(6)  p  =  a  —  6  sin  ^  ;  (cl)  p  =  a  sin  6. 

Ans.     (a)  p  =  a(l  +  sin^). 


138 


ANALYTIC    GEOMETRY 


108.   The   Polar   Equation  of  the    Conic.  —  If   the  pole  is 
taken  at  the  focus  and  the  directrix  is  perpendicular  to  the 
polar  axis  as  in  the  figure,  we  have  at  once  from  the  defini- 
tion of  a  conic,  OP  =  eMP, 
whence 

p  =  e(2^  +  p  cos  6), 

which  gives 


P  = 


ep 


(42) 


1  —  e  cos  6 

If  the  directrix  is  taken  par- 
allel to  the  polar  axis,  the 
equation  becomes 

ep 


9  = 


(42«) 


1  —  €  sin  6 

The  transformations  of  rotation  and  symmetry  give  eight 
forms  of  the  conic  equation  in  which  the  focus  is  at  the  pole 
and  the  directrix  perpendicular  or  parallel  to  the  polar  axis. 
They  are  :  _^  gp  .   g.-, 

n=- i ;     p  = =*= — i . 

l±ecos^  1  ±  e  sin  0 

In  the  first  set  the  principal  axis  is  the  polar  axis ;  in  the 
second  it  is  the  90°  axis.  To  sketch  a  conic  it  is  sufficient 
to  put  the  equation  in  standard  form  and  find  the  intercepts. 
The  standard  form  determines  the  principal  axis  and  eccen- 
tricity, and  the  intercepts  give  the  vertices  and  the  ends  of 
the  latus  rectum. 

By  using  the  polar  equation  many  of  the  properties  of 
the  conic  may  be  derived  more  easily  than  by  the  use  of  the 
rectangular  equation.  Some  of  these  are  indicated  in  the 
following  problems. 

PROBLEMS 

1.  Transform  the  equation  of  the  conic,  x^  -\-  y^  =  e-  {x  +  p)-,  into 
the  polar  form. 

2.  Show  that  the  conic  is  symmetrical  with  respect  to  the  principal 
axis.     Does  the  polar  form  show  any  other  axis  of  symmetry  ? 


POLAR  COORDINATES  139 

3.    Find  the  latus  rectum  of  the  conic.  Ans.   2  ep. 

4    Reduce  to  the  standard  polar  form  each  of  the  following  conies, 
find  e  and  p,  and  draw  the  curve  and  its  directrix : 

16  ,  -  5 

(«)  ^=2-cos^'  ^'^  ^  =  l  +  cos^' 

8  12 


1  -2  sin  0'  ''^  '^      4  +  cos^' 

(^^^  =  2  +  3'cos^'  (^)  ^  =  i_2cos^' 

Q 

Alls,     (a)  p  = e  =  1    n  =  16. 

^   -'  ^      1  —  ^cos^'  2'^ 

ep 

5.  Show  that  if  p  =- ;,is  the  equation  of  a  hyperbola,  the 

^      1  —  e  cos  d  .11 

inclinations  of  the  asymptotes  are  given  by  cos d  =  ±-. 

e 

6.  In  the  hyperbolas  of  Problem  4,  draw  the  asymptotes. 

7.  Find  the  polar  intercepts  of  the  conic.     From  these  show  that 

(a)  the  major  axis  of  the  ellipse  is  — ^^  ; 

(h)  the  transverse  axis  of  the  hyperbola  is  — ^  , 
^^  e--l 

8.  Show  that  the  conic  never  crosses  the  directrix. 

Hint.  —  The  equation  of  the  directrix  is  p  cosO  =  —  p.     Points  of  inter- 
section are  found  by  solving  simultaneously. 

9.  Derive  the  equation  of  the  conic  when  the  equation  of  the 
directrix  is : 

(a)  pcosd=p;  (b)  p?,'m.d=p. 

10.    Draw  the  conic  with  it-s  directrix  for  each  form  : 
^P        .  n,\  .-        ^P 


{^)  p-^^ — .;  i^)  p 


1  4-  e  cos  ^ '  1  +  e  sin  ^ ' 

ep  ep 

1  +  e  cos   d  1  +  e  sm  d 

ep  ep 

P  =  -. -'y  P~ 


P  = 


1  —  e  cos  d  1  —  e  sin  6 

^_P .  «__        ep 


1  —  e  cos  6  1  —  e  sin  6' 


11.    Show  that  the  locus  ot  p  =.  a  sec- -is  a  parabola. 


140 


ANALYTIC    GEOMETRY 


109.  Equations  derived  in  Polar  Coordinates. — The  equa- 
tion of  a  locus  may  sometimes  be  derived  more  readily  in 
polar  than  in  rectangular  coordinates.  This  is  especially 
true  when  the  locus  is  described  by  the  end  of  a  line  of 
varying  length  revolving  around  a  fixed  point.  The  process 
is  substantially  the  same  as  when  rectangular  coordinates 
are  used.  The  pole  should  be  chosen  at  a  convenient  place, 
usually  a  fixed  point  about  which  some  line  in  the  problem 
revolves.  Then  a  point  (p,  0)  satisfying  the  conditions  is 
chosen,  and  relations  between  p,  0  and  the  given  arbitrary 
constants  are  obtained  by  means  (usually)  of  trigonometric 
formulas. 

Example.  —  A  chord  through  the  end  of  a  diameter  of  a  circle  is 
extended  the  length  of  the  diameter.     Find  the  locus  of  the  end  of 

this  line. 

Choose  one  end  of  the  given  di- 
ameter as  the  pole,  and  the  diame- 
ter as  the  polar  axis.  Let  2  «  be 
the  diameter  of  the  given  circle. 

Then  from  the  figure  : 

OB  =  2  a  cos  e,  p  =  OB  +  2  a. 

Therefore  the  equation  is 

P  =  2  a(l  +  cos^), 

and  the  required  locus  is  a  cardioid. 


p(p.e) 


PROBLEMS 

1.  Find  the  locus  under  the  conditions  of  the  above  problem  when 
the  chord  is  extended  a  distance  2  b.  Ans.     The  limagon. 

2.  The  radius  of  a  circle  whose  center  is  the  origin  is  prolonged  a 
distance  equal  to  the  ordinate  of  its  extremity.  Find  the  locus  of  the 
end  of  this  line.  Ans.     The  cardioid,  p  =  a(l  +  sin  ^). 

3.  Derive  a  formula  for  the  distance  between  two  points  in  terms 
of  their  polar  coordinates. 

Hint.—  Use  the  law  of  cosines. 


POLAR  COORDINATES  141 

4.  Derive  the  polar  equation  of  the  circle  having  the  center  (pi.  ^i). 

Ans.     /)2  +  pi2  _  2  pi  p  cos  (d  -  ^i)  =  r^. 

5.  Derive  the  polar  equation  of  the  ellipse  when  the  pole  is  at  the 
center.  A^is.     p^(l  —  e^cos^d)  =  b'^. 

Hint.  —  Use  the  law  of  cosines  and  the  fact  that  the  sum  of  the  focal 
radii  is  2  a. 

6.  Find  the  locus  of  the  vertex  C  of  a  triangle  whose  base  AB  is 
fixed  and  which  has  the  angle  A  =  2  C. 

Ans.     p  =  a  (2  cos  ^  +  1),  the  trisectrix.* 

Hiiit.  — Use  the  law  of  sines. 

7.  A  fixed  point  A  is  at  a  distance  a  from  a  fixed  line  BC.  From 
^  a  line  is  drawn  cutting  BC  in  B  and  points  P  and  P'  are  chosen  on 
this  line  b  units  from  D  on  either  side.  Find  the  locus  of  P  and  P'  as 
D  slides  along  BC.  Ans.     p  =  a  sec  6  ±  b. 

8.  A  tangent  is  drawn  to  a  circle  whose  center  is  the  origin  and 
terminated  by  the  x-  and  ?/-axis.     Find  the  locus  of  its  mid-point. 

A71S.     p  sin  26  =  a. 

9.  Find  the  locus  of  the  mid-points  of  chords  drawn  from  the  end 
of  a  fixed  diameter  of  a  circle. 

10.  A  line  of  length  2  a  has  its  extremities  on  two  fixed  perpen- 
dicular lines.  Find  the  locus  of  the  foot  of  a  perpendicular  from  the 
intersection  of  the  fixed  lines  to  the  line  of  constant  length. 

Ans.     p  —  a  sin  2  6. 

11.  Given  the  circle  p  =  2  a  cos  6  and  the  line  p  cos  6  =  2  a.  From 
the  pole  a  chord  OB  is  drawn,  meeting  the  line  in  C.  Find  the  locus 
of  a  point  P  on  the  line  OC,  if  OP  always  equals  BC. 

Ans.     p  =2  a  tan 6  sin  6,  the  cissoid. 

12.  The  base  of  a  triangle  is  in  length  2  a.  Find  the  locus  of  the 
vertex  if  the  product  of  the  sides  is  equal  to  a-. 

Ans.     p^  =  2o2  cos  2  6.,  the  lemniscate. 

^m^  — Take  the  pole  at  the  center  of  the  base  and  use  the  law  of 
cosines. 

*  Since  the  vertical  angle  C  is  one  third  of  the  exterior  angle  at  B,  this 
form  of  the  lima^on  can  be  used  to  trisect  any  angle,  and  so  is  called  the 
trisectrix. 


142  ANALYTIC    GEOMETRY 

REVIEW   PROBLEMS 

Identify  and  sketch  the  following  curves  : 

1.  p  =  6sin6.  14.    p  cos  ^  =  —  5. 

2.  p=^     ^.      •  15.    p=        ^ 


4.  p  =  Sd. 

6.  p=10  sin  2  d. 

6.  p  =  6  tan  0  sin  ^. 

7.  p  =  4  -\-  6  sec  ^. 


8.    p 


11.    p  sin  6*  =10. 


2  -  sin 0  1  +  cos^ 

3.    pr=6  +  4sin0.  ig.   pzr5-10cos^. 


17.  p^  =  8  TT. 

18.  p2  =  4  cos  2  d. 

19.  p  =  8  cos  3  e. 
3                                       20.  p  =  1  +  sec  ^o 

1  +2COS0  21.  p2  =  4sin2  2^. 


9.    p  =  8cos2^.  22.   p=  -2cot6'csc9. 

10.    p  =  8-  10 sin 0. 


23.  2p  — pcos<9  =  8. 

24.  tan  ^=1. 


12.  p_4sec2-.  gg    p  =  10 sin 3  0. 

13.  p  =  6  sin  5  0.  26.    p  =  8  tan-  6 sec  0, 


^Wou^    Y^  ^  I.-/ -^  f--^    Jrl 


CHAPTER   IX 
HIGHER  PLANE   CURVES 

110.  Algebraic  and  Transcendental  Equations.  —  The 
equations  in  Cartesian  coordinates  which  we  have  hitherto 
treated  have  been  algebraic  equations,  i.e.  have  involved 
only  integral  and  fractional  powers  of  x  and  y. 

Any  equation  which  is  not  algebraic  (e.g.  y  +  sin  x  =  0) 
is  called  transcendental,  and  functions  defined  by  such  equa- 
tions are  called  transcendental  functions.  The  elementary 
equations  of  this  class  are  those  in  which  the  exponential, 
logarithmic,  trigonometric,  and  inverse  trigonometric  func- 
tions are  used. 

In  this  chapter  we  shall  discuss  a  number  of  curves  de- 
fined by  transcendental  equations  and  algebraic  equations  of 
degree  higher  than  the  second.  Such  curves  are  called 
Idgher  plane  curves. 

111.  The  Exponential  Curve.  —  This  is  defined  by  the  ex- 
ponential equation 

where  h  is  any  positive  constant.  The  quantity  h  is  called 
the  base.  If  the  exponent  is  fractional  and  involves  even 
roots  of  b,  only  the  positive  value  of  the  root  is  used. 

Discussion.  —  For  simplicity  consider  the  case  that  6  >  1. 

When  X  =  0,  y  =  b^  =  \.  If  ?/  =  0,  we  would  have  0  =  6^, 
which  is  impossible  for  any  value  of  x.  Therefore  the  curve 
crosses  the  ^/-axis  one  unit  above  the  origin  and  does  not 
cross  the  a;-axis. 

For  all  values  of  x,  positive  and  negative,  y  is  positive, 
since  any  power  of  a  positive  number  is  positive.     When 

143 


144 


ANALYTIC   GEOMETRY 


X  >  0,  b""  increases  with  x,  since  6  >  1.  Therefore  in  the 
first  quadrant  the  curve  recedes  to  infinity  away  from  both 

the  X-  and  y-Sixes.     Since  b'""  =  — ,  when  x  approaches  —  co, 

b"" 

y  approaches  0  as  a  limit.  Thus  in  the  second  quad- 
rant the  curve  approaches  the  a>axis  as  an  asymptote.  In 
the  following  figure  the  curve  is  plotted  for  two  values  of  b. 

The  Base  e.  —  The  most  important  case  of  the  exponential 
is  for  the  base  e(=  2.718+),  the  base  of  the  natural  system 
of  logarithms. 

In  the  figure  the  dotted  line  is  the  curve  oi  y  =  e""  and  the 
heavy  line  that  of  y  =  2'. 


X 

2/=«^ 

y  =  '2' 

-2 

0.135 

0.250 

1 

0.368 

0.500 

0 

1. 

1. 

1 

2.718 

2. 

2 

7.389 

4. 

3 

20.086 

8. 

Exercise  1.     Discuss  the  effect  upon  the  graph  of  increasing  b. 

Exercise  2.     Discuss    the    exponential   equation   for :  (a)    6  <  1  ; 
(6)  b  =  1. 


112.   The  Logarithmic  Curve 
equation 


This  is  the  graph  of  the 


y  =  log^  X. 

The  base  b   is  always  positive  and  different  from  1,  and 
usually  greater  than  1. 

Since  in  Algebra  the  logarithm  of  a  number  to  a  given 
base  is  defined  as  that  power  of  the  base  which  makes  the 
result  equal  to  the  given  number,  the  equation  y  =  loQf^x  may 
be  written  in  the  form  x  =  b".  This  is  the  exponential  equa 
tion  with  the  variables  interchanged.     Thus  the  logarithmic 


HIGHER  PLANE   CURVES 


145 


equation  is  obtained  from  the  exponential  y  =  5^  by  solving 
for  X  and  interchanging  variables.  Logarithmic  and  expo- 
nential functions  are  said  to  be  the  inverse  of  each  other. 

The  discussion  of  the  logarithmic  equation  y  =  log^  x  will 
therefore  follow  from  that  of  the  exponential  equation  y  =  h'^ 
on  interchanging  variables.  Thus  the  logarithmic  curve 
crosses  the  avaxis  at  (1,  0)  and  does  not  cross  the  ?/-axis. 
For  .r  >  1,  2/  >  0  and  as  x  =*co,  y  =  cc  ;  for  x  <  1,  y  <  0  and 
as  x  =  0,  2/  =  —  GO.  Since  x  =  b",  the  abscissa  x  is  positive 
for  all  real  values  of  y,  whether  positive  or  negative.  This  is 
the  reason  for  the  statement  that  "  logarithms  of  negative 
numbers  do  not  exist." 

The  following  graph  is  the  logarithmic  curve  for  b  =  10. 
The  table  of  values  may  be  computed  by  using  a  table  of 
logarithms  or  by  writing  the  equation  in  the  form  x  =  10^ 
and  computing  values  of  x  corresponding  to  various  values 
of  y.  Note  that  the  logarithmic  curve  is  symmetrical  to  the. 
exponential  curve  with  respect  to  the  line  y  =  x. 


y  = 

logio  X 

X 

y 

X 

y 

1.0 

0 

.1 

-1 

3.1 

.5 

.01 

-2 

10 

1 

.001 

-3 

100 

2 

.0001 

-4 

etc. 

etc. 

etc. 

etc. 

113.  Applications.  —  The  exponential  and  logarithmic 
curves  show  clearly  the  method  of  variation  of  these  func- 
tions. Thus  for  &  >  1,  the  exponential  function  increases 
with  great  rapidity  as  x  approaches  go,  while  the  logarithmic 
function  increases  very  slowly. 


This  symbol  is  used  for  the  words  "  approaches  as  a  limit." 


146  ANALYTIC   GEOMETRY 

These  functions  not  only  have  a  theoretical  importance, 
but  constantly  appear  in  the  statements  of  the  laws  of  Phys- 
ics and  other  sciences.  For  example,  the  adjoining  figure 
represents  a  weight  W  suspended  by  a 
/^     "'N^  rope  wrapped  several   times    about    a 

wooden  beam  and  kept  from  falling  by 
a  tension  T.  The  relation  between  W 
and  T  is  given  by  the  equation 

where  x  represents  the  number  of  times 
the  rope  is  wound  around  the  beam  and  /x  is  a  constant  de- 
pending on  the  friction  between  the  rope  and  the  beam. 
For  a  hemp  rope  on  smooth  oak  /x  =  3.34,  nearly. 

PROBLEMS 

1.  Plot  the  system  of  curves  defined  by  ?/  =  &^  for  6  =  1,  2,  d,  and 
10 ;  all  on  the  same  axes. 

2.  Plot  the  system  defined  hj  y  —  h'  ior  b  =  1,  ^  and  \ ;  all  on  the 
same  axes. 

3.  Plot  the  logarithmic  curve  for  the  bases  10,  3,  e,  and  2 ;  all  on 
the  same  axes. 

4.  Plot  the  graphs  of : 

(«)  y  =  2-^;       (c)  y  =  x^^;       (e)  y  =  e-^  ; 

(b)  y=4-';      (d)  y  =  Q-;     (/)  y  =  e^-. 

5.  Plot  the  graphs  of  : 

(a)  y  =  Logo  X  ;  (c)  2x  =  logio y  ; 

(6)  y  =  S\og2X;  (d)     x  =  logio2^ 

6.  Write  each  of  the  equations  of  Problem  4  in  logarithmic  form. 

Ans.    (a)  X  =  —  logo  y. 

7.  Write  each  of  the  equations  of  Problem  5  in  exponential  form. 

Ans.    (a)  X  =  2y. 

8.  Discuss  the   following  equations  for  symmetry  and  piot  their 
graphs : 

pX    _|_    p~X 

{(^)  y  =  — .    This  is  called  the  hyperbolic  cosine  and  written 

y  —  Qosh  X. 


HIGHER  PLANE   CURVES 


147 


oX   __    P~X 

(b)  y  = ,  or  y  =  sink  x. 


(c)  y=-  (e"  +  e""),  the  equation  of  the  catenary.* 


9.    Show  that  the  graphs  of  y  =  b""  and  y  =  (-\    are  symmetrical 
with  respect  to  the  ?/-axis. 

10.  Plot  the  curve  i/  =  3^  and  estimate  the  logarithms  to  the  base  3 
of  the  integers  from  2  to  10. 

11.  Plot  the  graph  of  the  function  W  =  Te^^,  where  T  =  100  lbs.  and 
fj.  =  l.l.  Estimate  from  the  graph  how  many  turns  of  the  rope  would 
be  required  for  this  tension  to  support  1  ton.    2  tons.    10  tons. 

Hi7it.  —  This  makes  ei^  =  3,  nearly. 


114.  Periodic  Functions.  —  A i^eriodic  function  of  a  variable 
is  a  function  ichose  values  are  repeated  at  definite  intervcds  as 
the  variable  increases.  Cosine  x  is  an  example.  It  passes, 
successively  through  all  values  from  +1  to  —1  and  back  to  + 1 
as  its  angle  increases  from  0  to  2  tt,  and  repeats  these  values  in 
the  same  order  as  often  as  the  angle  x  increases  by  2  tt  ;  or  n 
times  for  a  change  of  2  mr  in  the  angle. 

The  interval  of  repetition  is  called 
the  period  of  the  function.  If  k  is 
the  period  of  f(x),  then  f{x-\-'k) 
=  f(x).  The  maximum  numerical 
value  of  the  function  is  called  the 
amplitude.  The  table  illustrates 
these  definitions  for  several  func- 
tions. 


/(^•) 

Period 

Ampli- 
tude 

Sina: 

27r 

1 

Cosx 

27r 

1 

Tana: 

TT 

CO 

Sin  2  X 

TT 

1 

Cos- 

2  IITT 

1 

n 

115.  The  Sine  Curve.  —  This  is  plotted  from  the  equation 
y  =  sin  x,  where  x  is  to  be  reckoned  in  circular,  or  radian 
measure. 


*This  is  the  form  assumed  by  a  perfectly  flexible  cord  or  chain  sus- 
pended between  two  points. 


148 


ANALYTIC   GEOMETRY 


We  may  assume  values  for  x  at  any  convenient  interval, 

as  ^ ,  and  take  the  corresponding  values  of  sin  x  from  a  table 

of  natural  sines. 

In  the  accompanying 
table  the  values  of  x 
in  degrees  are  found  in 
column  2  and  their  cir- 
cular measures  to  be 
used  in  plotting  are  in 
colunm  1.  The  circu- 
lar measure  of  x  is  the 
abscissa  and  the  value 
of  its  sine  is  the  or- 
dinate. 

In  plotting  it  is  con- 
venient to  lay  off  on 

the  .T-axis  —  =  .52  units 
6 

and,    using   this   as    a 

unit,  to  mark  off  the 

various     abscissas. 

After   the   values   from    the    table    up   to   2  tt  have   been 

plotted  the  curve  is  constructed  through  one  period,  and  the 


X 

(radians) 

X 

(degrees) 

sin  x 

cos  X 

0 

0 

.00 

1.00 

^=  .52 
6 

30 

.50 

.87 

1.05 

60 

.87 

.50 

1  =  1-^^ 

90 

1.00 

.00 

2.09 

120 

.87 

-.50 

2.62 

150 

.50 

-.87 

7r  =  3.14 

180 

.00 

-1.00 

3.66 

210 

-  .50 

-.87 

4.19 

240 

-.87 

-  .50 

4.71 

270 

-  1.00 

.00 

5.24 

300 

-.87 

.50 

5.76 

330 

-  .50 

.87 

2  TT  =  6.28 

360 

.00 

1.00 

6.81 

390 

.50 

.87 

7.33 

420 

.87 

.50 

etc. 

etc. 

etc. 

etc. 

same  values  of  y  may  be  used  to  construct  it  through  suc- 
cessive periods. 

Since  the  angle  x  may  be  taken  negative  as  well  as  positive, 
the  curve  may  be  extended  indefinitely  in  the  negative  direc- 
tion. If  we  substitute  {—  x,  —  y)  for  {x,  y)  in  the  equation, 
we  nave     —  y  =  sin  (—  a;)  =  —  sin  x,  or  y  =  sin  x^ 


HIGHER  PLANE   CURVES 


149 


and  the  form  of  the  equation   is   unchanged.     Hence   the 
curve  is  symmetrical  with  respect  to  the  origin. 


116.  Circular  Measure.  —  For  plotting  the  sine  curve  x 
can  be  measured  in  degrees,  and  any  convenient  unit  taken 
as  a  degree,  but  the  scale  of  measurement  in  x  will  not  corre- 
spond to  that  in  y  unless  circular  measure  is  used.  Circular 
measure  gives  the  same  scale  for 
both  coordinates.     For  in  radians, 

arc 


angle  = 


or 


X 


radius ' 
AB 


But  y  =  sin  x  = 


r 
AC 


T 


So  as  the  angle  x  is  generated 
by  rotating  the  side  OA  from  the 
initial  position  OB,  the  lengths  of  AB  and  AC  in  terms  of 
the  radius  as  a  unit  are  the  values  of  x  and  sin  x  given  in 
columns  1  and  3  of  the  table  of  values  for  plotting.  This 
is  only  true  where  x  is  expressed  in  circular  measure. 

117.  The  Cosine  Curve.  —  The  curve  of  cosines  is  plotted 
from  the  equation  y  =  cos  x.  This  curve  has  the  same 
period  and  amj)litude  as  the  sine  curve,  and  in  fact  has  the 
same  shape ;  but  it  passes  through  (0,  1)  instead  of  the 
origin. 

This  is  true  because  the  cosine  of  any  angle  is  equal  to 
the  sine  of  an  angle  90°  greater,  or  cos  x  =  sin  (90°  +  x). 
Hence  the  table  of  sines  can  be  used  for  plotting  the  cosine 
curve  by  moving  the  whole  column  of  sine  values  up  90°,  so 
that  the  sine  curve  becomes  the  cosine  curve  when  the  origin 


is  advanced  to  the  point  [  ^,  OV 


150 


ANALYTIC    GEOMETRY 


118.  The  Tangent  Curve.  —  The  curve  of  tangents  is 
plotted  from  the  equation  y  =  tan  x.  It  passes  through  the 
origin  and  its  period  is  tt.     The  value  of  y  is  infinite  when 


TT 


X  is  an  odd  multiple  of  -,  and  the  curve  passes  through  all 

values  of  v  between  a:  =  —  -  and  a;  =  - ,  and  so  on  at  inter- 

2  2 

vals  of  TT,  the  period. 

The  table  of  values  follows. 


X 

tan  X 

IT 

—  oo 

2 

TT 

~3 

-1.73 

■K 

~  6 

-.58 

0 

0. 

TT 

.58 

TT 

3 

1.73 

W 

2 

x 

119.    Multiple   Angles.  —  Consider   y=?>mnx.     The   mul- 
tiple n  cUcides  the  period  of  the  function  by  n,  but  does  not 

alter  the  anqjlitude.     To  prove  this,  substitute  x -\-  "^  f or  x. 

n 

We  have       •       /     ,  2  ttX        •     ^       ,   o    \ 

sm  n    X  -\ =  sm  (nx  +  J  tt)  =  sm  nx. 


Hence  -^  is  the  period  of  sin  nx.     That  the  amplitude  i? 
n 

unaltered  is  obvious. 


HIGHER  PLANE   CURVES 


151 


Compare  this  with  y  =  n  sin  x,  where  the  period  is  un- 
altered, but  the  amplitude  is  multiplied  by  n.  All  of  this 
appears  clearly  in  the  plots.     Thus  we  have 

for  y  =  sin  x,     the  period  is  2  tt  and  y  =  1,  when  x  =  90°  ; 

for  ?/  =  sin  2  x,  the  period  is      w  and  y  =  1,  when  x  =  45° ; 

for  y  =  2  sin  x,  the  period  is  2  tt  and  y  =  2,  when  x  =  90°. 


120.  Sum  of  Functions.  —  Sometimes  it  is  necessary  to 
plot  functions  which  are  the  sum  of  two  or  more  trigono- 
metric functions.  Time  is  saved  in  making  the  table  of 
values  by  writing  the  values  of  the  various  terms  in  ]3arallel 
columns   and   summing    corresponding   values    to   get   the 


various  ordinates.  Thus  to  plot  y  =  sin  2  a;  +  2  cos  x, 
make  a  table  of  values  for  sin  2  x  and  2  cos  x  in  parallel 
columns  and  write  the  sum  of  values  corresponding  to  the 
same  value  of  x  in  a  third  column  for  the  ordinate  to  be 
plotted. 

An  alternative  method  is  to  plot  the  curves  y'  =  sin  2  x  and 
y"  =  2  cos  X  on  the  same  axes  and  obtain  the  correspond- 


ing ordinates  by  measurement, 
method. 


The  figure  illustrates  this 


152 


ANALYTIC    GEOMETRY 


121.  Inverse  Trigonometric  Functions.  —  The  relation  ex- 
pressed in  the  inverse  notation  by  2/  =  sin~^  x  reads  in  direct 
notation  x  =  sin  y. 

y  =  cos~i  X  =  arc  cos  x. 


y  =  sin~i  X  =  arc  sin  x. 
Y 


The  curve  of  inverse  sines  is  therefore  the  same  as  the 
curve  of  sines  in  shape,  period,  and  amplitude,  but  the  axes 
are  interchanged. 

The  notation  arc  sin  x  is  preferable  to  sin~^  x. 


PROBLEMS 

1.  Discuss  and  plot  the  graph  of  : 

(a)  y  =  cos  X  ;  (c)  y  =  sec  x  ; 

(6)  y  =  cot  x;  (d)  y  =  esc  x. 

2.  Find  the  period  and  amplitude  and  draw  the  graph  of 
(a)  y  =  s'mSx;  (d)  y  =  SsecSx; 
lb)  y  =  cosbx;                                    (e)  ?/  =  |cos|x. 

(c)  y  =  tan  -. 


3.    Plot  the  graph  of 

(a)  y  =  arc  tan  x ; 

(b)  y  =  arc  sec  x  ; 


(c)  y  =  2  arc  cos  2x: 
3  arc  sin  - . 


id)  y 

o 

4.   Draw  on  the  same  axes  the  graphs  of  : 

raW^^^'^^'  (b)    \y  =  ^^^^^  f(.^    f^  =  tanx, 

^  ^    \^y  =  CSC  x  ;  ^   ^    \^y  =  sec  x;  ^^    ]y  =  cot  x. 

What  relation  holds  between  corresponding  ordinates  of  each  pair 


of  curves  ? 


V^j 


(yjJjO^A 


> 


/^yjb^  3  /-    - 


HIGHER  PLANE   CURVES  153 

5.  Draw  each  of  the  following  curves.  For  any  value  of  ar,  draw 
ordinates  at  ±  ^,  tt  ±  a*,  2  tt  ±  a:,  and  then  write  the  trigonometric  for- 
mulas suggested  by  the  figure. 

(a)  2/  =  sinx;  (c)  ?/ =  tan  x  ;  (e-)  «/ =  sec  x  ; 
•  (6)  y  =  cos  %\                  ((Z)  y  =  cot  x  ;                  (/)  2/  =  esc  x. 

Ans.    (a)  sin  (— .r)  =  — sinx  ; 

sin  i^TT  ±  x)  =  ^  sin  x  ; 

sin  (2  TT  ±  ar)  =  ±  sin  X. 

6.  Plot  the  graphs  of  the  following  equations  : 

(rt)  y  =  cos  2  a:  —  sin -;  (c)   ?/ =  a:  +  2  sin  a; ; 

(b)  y  =zsecx  —  x;  (d)  y  =  cos|  +  sin  2 x. 

7.  Determine  which  of  the  functions  of  Problem  6  are  periodic  and 
find  the  periods. 

Alls,  (a)  and  (cZ)  have  the  period  4  tt  ;  (6)  and  (c)  are  not  periodic. 

122.  Parametric  Equations.  —  In  some  curves  the  coor- 
dinates are  functions  of  a  third  variable  such  that  the  condi- 
tions determining  the  curve  may  conveniently  be  expressed 
by  two  equations  between  these  three  variables  instead  of  by 
a  single  equation  in  x  and  y.  This  third  variable  is  called  a 
parameter,  and  the  two  equations  jjarametric  equations  of  a 
curve.  If  t  is  the  parameter,  the  parametric  equations  of  the 
curve  are  usually  of  the  form  x  =f(t),  y  =  g{t). 

The  parameter  usually  represents  some  geometric  mag- 
nitude, or  the  time  during  which  the  point  tracing  the  curve 
has  been  in  motion  ;  but  it  may  be  chosen  in  any  manner  that 
is  convenient.  Thus  for  the  semi-cubical  parabola  y"^  =  a^^ 
we  may  assume  a  parameter  t  and  write  x  =  at^,  y  =  aH^,  since 
by  eliminating  t  between  these  equations  we  obtain  the 
original  form. 

To  plot  the  graph  of  such  equations  we  take  various 
values  of  the  parameter  and  compute  the  corresponding 
values  of  x  and  y.  Then  plot  the  points  {x,  y)  as  usual. 
The  values  of  the  parameter  appear  only  in  the  table  of 
values,  and  not  in  locating  the  point. 


154 


ANALYTIC   GEOMETRY 


123.  Parametric  Equations  of  the  Circle.  —  As  usual  wo 
take  the  origin  at  the  center  of  the  circle.  Then  the  circle 
is  generated  by  rotating  OP.  Taking  6  as  the  parameter, 
we  have  at  once  from  the  figure  and  the  definitions  of  the 
sine  and  cosine  : 

X  =  r  cos  B,         y  =  r  sin  6. 

These  are  the  parametric  equations  of  the  circle.     Erom  the 

table  of  values  below  points  on  the  curve  are  (10,  0),  (8.7,  5), 

(5,  8.7),  (0,  10),  (-5,  8.7),  (-  8.7,  5),  (-  10,  0),  (-  8.7,  -  5), 

etc. 

Table  of  Values  for  r  =  10 


P{x,y) 


9 

X 

y 

0 

10 

0 

30 

8.7 

5. 

60 

5. 

8.7 

90 

0 

10 

120 

-5. 

8.7 

150 

-8.7 

5. 

180 

-  10. 

0. 

210 

-8.7 

-  5 

Note  that  if  we  eliminate  0  between  the  parametric  equa- 
tions, we  get  the  standard  form  x"^  -\-  if  =  r^. 

124.  Parametric  Equations  of  the  Ellipse.  —  Suppose  we 
have  a  circle  of  radius  a  and  center  0,  in  which  OA  is  any 
radius  and  AMsl  perpendicular 
from  A  upon  a  fixed  diameter. 
5  is  a  point  on  the  radius  at 
a  distance  b  from  0  and  P  is 
the  projection  of  B  on  AM. 
It  is  required  to  find  the  locus 
of  P 

Take  the  fixed  diameter  as 
the  X-axis  and  the  center  as  the 


HIGHER  PLANE   CURVES 


loo 


origin.     Let  (x,  y)  be  the  coordinates  of  P  and  6  =  /.MO A 
be  the  parameter.     Then  we  have  at  once  from  the  figure, 

X  =  OM  =  OA  cos  ^  —  a  cos  0, 
y  =  MP=  OB  sin  (9=  6  sin  0. 

These  are  the  parametric  equations  of  the  locus.  It  remains 
to  show  that  the  curve  is  an  ellipse.  Solving  the  equations 
for  cos  6  and  sin  6  respectively,  squaring  and  adding  we  get 


«2  ^  b^ 


cos2  ^  +  sin2  ^  =  1, 


the  equation  of  the  ellipse. 

The  circles  described  by  B  and  A  are  called  the  auxiliary 
circles  of  the  ellipse.  By  drawing  them  on  coordinate  paper, 
the  various  positions  of  P  can  be  readily  found  and  any 
ellipse  of  given  semi-axes  plotted  with  accuracy. 

125.  Path  of  a  Projectile.  —  If  an  object  moves  with  a 
velocity  of  V  feet  per  second  along  a  line  inclined  to  the  x- 
axis  at  an  angle  a,  it  is  clear 
from  the  adjoining  figure  that 
at  the  end  of  each  second  its 
distance  from  the  ?/-axis  has 
been  increased  b}'  V^=  V  cos  a 
feet,  and  that  from  the  .r-axis 
by  Vy  =  V  sin  a  feet.  These 
quantities,  V^  and  V^,,  are  called 
the  X-  and  y-components,  respec- 
tively, of  the  velocity,  and  when 
multiplied  by  the  time  give  the  total  displacement  parallel 
to  the  X-  and  y-axes. 

In  the  case  when  the  motion  is  in  a  vertical  plane,  the 
vertical  velocity  does  not  remain  constant,  but  is  affected 
by  the  force  of  gravity.  In  this  case  we  find  in  Mechanics 
that  the  height  is  given  by  the  formula  h  =  vt—  16  f^,  where 


r 

^ 

Vy^ 

yy 

y 

y. 

0 

X 

156  ANALYTIC    GEOMETRY 

li  is  the  height  in  feet,  t  the  time  in  seconds,  and  v  the  ver- 
tical  component  of  the  initial  velocity  in  feet x>er  second* 

Using  these  principles,  let  ns  solve  the  problem  :  Find 
the.  path  of  a  projectile  discharged  at  an  angle  «=  arc  cos  -| 
with  an  initial  velocity  of  80  feet  per  second. 

Taking  the  point  of  projection  as  the  origin,  the  ?/-axis  vertical  to 
the  earth's  surface  and  the  x-axis  horizontal  in  the  direction  of  projec- 
tion,  let  F(x,  y)  be  the  position  of  the  projectile  after  t  seconds. 

The  horizontal  component  of  the  initial  velocity  is  Vx  =  80  cos  a  = 
80  •  f  =  48  ;  the  vertical  component  is  F^  =  80  sin  a  =  80  •  f  =  64.     As 

the  horizontal  velocity  remains  con- 
stant, X  =  Vxt  =  48t.  (a) 
Using  the  above  formula, 
y  =  Vyt  -  \6  f^  z=  Q4:  t  -  16  f^  (b) 
Equations  (a)  and  (h)  are  the 
parametric  equations  of  the  locus. 
To  identify  the  curve,  eliminate  t 
and  we  have 

?/  =  gX  —  xfi^  ? 
evidently  a  parabola. 
To  get  the  range,  set  y  =  0,  and  we  have  x  =  192  feet.     If  we  com- 
plete the  square,  we  have 

(X- 96)2=  -144(2/ -64). 
Hence  the  vertex  is  (96,  64),  i.e.  the  greatest  height  reached  is  64  feet. 

PROBLEMS 

1.  Plot  the  graphs  of  the  following  parametric  equations : 

1  t^ 

(a)  x  =  St,y  =  —;  ,(e)  x  =  ti-l,y=-; 

(b)  X  =  t"^,  y  =  2  t  ;  (f)  x  =  acosd,  y  =  b sinS  0  . 

(c)  X  =  5  cos  6,  ?/  =  3  sin  ^  ;         (g)  x  cos  d  =  a,  y  =  a  tan  d  ; 

(d)  x  =  a  cos'^^,  y  =  bsin^d]     (h)  x  =  2  +  3cos  ^,  ?/  =  3sin^  -  1. 

2.  Eliminate  the  parameter  in  each  pair  of  equations  in  Problem  1. 

3.  Form  parametric  equations  for  the  folium  of  Descartes  x^  +  y^ 
=  3  axy,  make  a  table  of  values  and  plot  the  curve. 

Hint.  —  Set  y  =  tx.  ,  3 ««  3  af^ 

*In  this  and  all  that  follows,  we  disregard  air  resistance.  The  results 
are  therefore  first  approximations  to  the  true  ones. 


HIGHER  PLANE   CURVES  157 

4.  A  straight  line  has  the  a;-intercept  5  and  the  inclination  30^. 
Find  its  parametric  equations  when  the  parameter  t  is  the  distance  of 
the  tracing  point  from  the  intersection  with  the  a:-axis. 

Ans.     x  =  b+  «— ,  y  =Z' 

6.  Find  the  locus  of  the  projectile,  the  range,  and  highest  point 
reached  from  the  following  data : 

(«)  initial  velocity  576  feet  per  second,  angle  of  projection  45°; 
(6)  initial  velocity  400  feet  per  second,  angle  of  projection  30°  ; 
(c)  initial  velocity  800  feet  per  second,  angle  of  projection  45°. 

6.  In  Problem  5  (a),  the  projectile  is  fired  up  a  hill  of  slope  ^. 
Find  how  far  up  the  hill  it  will  strike  the  ground. 

Ans.     5796  feet. 

7.  Find  the  equation  of  the  locus  of  a  projectile  of  initial  velocity 

16.7:2 

Fand  angle  of  projection  a.  Ans.     y  =  a:  tan  a  — 


F^cos^a 


8.    In  Problem  7  find  the  range  and  greatest  height. 


Ans.     Range — — ■  ;  greatest  height — . 

o2  o4 

9.  A  fly  crawls  on  a  spoke  of  a  wheel  from  the  hub  toward  the 
rim  at  a  rate  of  4  feet  per  minute.  The  wheel  makes  2  revolutions  per 
minute.     Find  the  equation  of  the  path  described  by  the  fly. 

A}is.     (in  polar  coordinates)  p  =  -  • 

IT 

10.  A  point  moves  around  a  circle  of  radius  10  feet  50  times  per 
minute.  Write  the  parametric  equations  of  the  circle,  using  the  time 
during  which  the  point  has  been  in  motion  as  a  parameter. 

11.  The  parametric  equations  of  the  path  of  a  point  are 

X  =  a  cos  kt, 

y  =  o, 

where  x  is  measured  in  feet,  t  is  measured  in  seconds  and  a  and  k  are 
constants.  Plot  the  locus.  How  often  does  the  point  pass  through 
the  origin  ? 

126.  The  Cycloid.  —  If  a  circle  rolls  along  a  straight  line 
the  curve  traced  by  a  point  on  the  circumference  is  called 
a  cycloid.     The  straight  line  is  called  the  base. 


158 


ANALYTIC   GEOMETRY 


As  the  circle  rolls  it  describes  the  length  of  the  circum- 
ference on  the  base  at  each  revolution,  and  a  proportional 

distance  for  any  part  of  a 
revolution.  Hence  the  point 
of  the  circle  touching  the 
base  is  always  at  a  distance 
from  the  starting  point  equal 
to  the  length  of  the  arc  sub- 
tending 6. 

In  the  figure  the  tracing 
point  P  starts  from  the  point  of  contact  0,  which  is  taken 
as  the  origin,  and  the  base  is  taken  as  the  a>-axis. 

Call  the  angle  which  the  radius  to  the  tracing  point 
makes  with  the  vertical,  6.  Then  OA  =  PA  =  rO,  since 
arc  =  angle  x  radius.     This  gives 

x=OA-BA  =  rO  -  r  sin  0, 

y  =  AC-  DC=  r-r  cos  6, 

which  are  the  parametric  equations  of  the  cycloid. 

llie  Prolate  and  Curtate  Cycloids.  If  the  tracing  point 
is  taken  anywhere  on  the  radius  of  the  rolling  circle  or  the 
radius  produced,  the  curve 
is  called  the  x>rolate  cycloid 
when  P  is  without  the  circle 
and  the  curtate  cycloid  when 
P  is  within.  The  term 
trochoid  is  also  used  for  both 
prolate  and  curtate  cycloids. 

Let  a  be  the  distance  from 
the  center  to  the  tracing  point.     The  parametic  equations 
of  both  prolate  and  curtate  cycloids  are 

X  =  rO  —  a  sin  0, 
y  =  r—  a  cos  0. 


HIGHER  PLANE   CURVES  159 

They  are  derived  in  precisely  the  same  manner  as  those 
of  the  ordinary  cycloid. 

PROBLEMS 

1.  Sketch  the  path  of  a  point  on  a  28-inch  bicycle  wheel  for  two 
revolutions.  Find  the  base  of  the  cycloid  and  the  coordinates  of  the 
tracing  point  when  the  wheel  has  made  one  third,  one  half  and  two 
thirds  of  a  revolution. 

2.  Plot  the  cycloid  for  r  =  10,  forming  a  table  of  values  for  ^, 
X,  and  y. 

3.  Eliminate  6  between  the  parametric  equations  of  the  cycloid. 
Hi7it.  —  6=  arc  cos  (^      ^  ]  and  vers  ^  =  1  —  cos  6, 


Ans.     x  =  r  arc  vers  "  —  V2  ry  —  y'^. 
r 

4.  Write  out  in  full  the  derivation  of  the  parametric  equations  of 
(a)  the  curtate  cycloid  ;  (6)  the  prolate  cycloid. 

5.  Trace  the  curtate  cycloid  though  two  periods  when  r  =  6  and, 
a  =  5,  and  find  the  coordinates  of  the  tracing  point  when  6  =  120^. 

6.  Plot  the  prolate  cycloid  for  r  —  lO  and  a  =  12. 

7.  Show  from  the  equations  of  the  cycloid  that  y  is  a,  periodic 
function  of  x,  of  period  2  ttv. 

8.  Construct  the  figure  and  derive  the  equations  of  the  cycloid 
when  d  >  90°. 

9.  Show  that  the  first  arch  of  the  cycloid  is  symmetrical  with  respect 
to  the  line  x  =  irr. 

Hint.  —  Show  that  for  any  value  of  ?/,  as  2/1,  6  =  di,  ov  2'ir  —  di.  From 
this  get  two  values  of  x,  Xi  and  x.2,  and  show  that  x.i~  tt  r  =  tt  r  —  Xp  Then 
apply  the  definition  of  symmetry. 

127.  The  Epicycloid  and  Hypocycloid.  —  When  the  base 
is  a  circle  instead  of  a  straight  line,  the  curve  is  called  an 
epicycloid  if  the  generating  circle  rolls  on  the  outside  of 
the  base,  and  a  hypocycloid  if  it  rolls  inside. 

Note.  —  The  epicycloid  is  the  basis  of  the  "  Theory  of  Epicycles  "  which 
was  used  in  the  Ptolemaic  system  of  astronomy  to  explain  the  motions  of 
the  planets. 


160 


ANALYTIC   GEOMETRY 


Let  the  tracing  point  start  from  A  and  the  generating 
circle  move  on  the  outside  until  its  contact  with  the  base  is 

at  D.  Thus  the  arc  AP  of 
an  epicycloid  is  described 
and  the  arc  PD  has  rolled 
over  the  arc  AD  of  the  base. 
Let  the  angle  which  the 
generating  radius  CP  makes 
with  the  line  of  centers  be  <^, 
and  the  angle  of  this  line  CO 
with  the  ic-axis  be  6. 

Using     circular     measure, 
since  arc  AD  =  arc  PD, 


BO  =  r<l>. 
Now  CPN  is  the  supplement  oi  0  +  cfi, 


and 


But 


and 


sin  CPN  =  sin  (0  +  cf>), 
cos  CPX=  -  cos  {0  +  cf>). 

x=  OB=OE  +  XP 

=  (R  +  r)  cos  $  —  r  cos  (6  +  <^) 
y=BP=EC-NC 
=  {R-\-  r)  sin  O  —  r  sin  (^  +  <^). 


(a) 


These  equations  contain  two  parameters,  6  and  (f>.     cf>  may 
be  eliminated  by  substituting  its  value  from  (a),  which  gives 

x  =  (i?  +  ?')  cos  O—r  cos  — ^  Of 

r 

y  =(^R  -\-r)  sin  O  —  r  sin  — ^t_  $^ 


HIGHER  PLANE   CURVES 


161 


If  r  is  taken  negative  we  have  the  equations  of  the  hy- 
pocycloid : 

a;  =  (i2  -  r)  cos  ^  +  r  cos  ?^^^  0,  ^ 

r 

y  ==(E  -  r)  sine -r  sin  ^^^  6. 

r 

The  Astroid.  —  In  the  hypo- 
cycloid  B  =  4:1',  the  rolling  circle 
makes  four  revolutions  in  pass- 
ing around  the  base,  forming  the 
astroid,  or  hypocycloid  of  four 
cusps. 

128.  The  Involute  of  the  Circle.  —  A  string  is  wound  about 
the  circumference  of  a  circle.     One  end  is  fastened  to  the 

circumference  and  the  string 
is  unwound.  If  the  string 
is  kept  stretched  the  curve 
traced  by  the  free  end  is  called 
the  involute  of  the  circle. 

If  it  begins  to  unwind  at 
A,  the  arc  AP  of  the  involute 
is  traced  when  the  string  has 
unwound  as  far  as  B,  and  the 
part  unwound  is  BP. 

BP  is  tangent  to  the  circle 
at  B,  and 


The  equations  are 


BP=BA  =  Be. 


x  =  R  cos  0  +  Bd  sin  9, 
y  =  EsiiiO-Pe  cos  0. 

Note.  — The  teeth  of  large-geared  wheels  are  cut  in  the  shape  of  epi-  or 
hypocycloids  as  the  best  forms  to  avoid  the  binding  and  friction  that  occur 
between  straight  teeth. 


162 


ANALYTIC    GEOMETRY 


PROBLEMS 

1.  Show  that  for  the  astroid  the  equations  reduce  to 

X  =  B  cos-^  6, 
y  =  B  sin3  e, 

2  2_  2. 

whence  :r^  +  y^  =  B^. 

2.  Simplify  the  equations  of  the  epicycloid  and  hypocycloid,  and 
draw  the  curves  when 

{a)  B  =  r:  (h)  B  =  2  r. 

3.  Show  that  the  epicycloid  in  which  i?  =  ?-  is  a  cardioid. 

4.  Draw  the  epicycloid  and  hypocycloid  when 

(a)  4i?  =  3r:  (6)  SB  =  or]  (c)  r  =  2B. 

5.  Derive  the  equations  of  the  hypocycloid  from  a  figure. 

129.  Algebraic  Curves. — Algebraic  curves  of  degree  higher 
than  the  second  have  a  great  variety  of  forms  and  are  best 
studied  with  the  aid  of  Calculus.  Two  simple  examples  of 
these  curves  are  the  cubical  x^arabola  y  =  a:i^  and  the  semi- 
cubical  parabola  y-  =  ax^.  The  majority  of  the  curves  in 
Chapter  YIII  are  found  to  be  algebraic  curves  if  their  equa- 
tions are  transformed  into  rectangular  form.  Many  of  these 
curves  are  of  historical  importance,  as  the  limacon,  the  con- 
choid, the  cissoid,  and  the  lemniscate,  given  in  Chapter  YIII, 
and  the  witch,  the  strophoid,  and  the  ovals  of  Cassini,  which 
will  be  discussed  in  the  following  sections. 

130.  The  Witch.  —  A  circle  of  radius  a  is  drawn  tangent 
to  the  :/>axis  through  the  origin,  meeting  the  y-axis  in  A. 

Y 


HIGHER  PLANE    CURVES 


163 


Through  A  a  tangent  is  drawn  to  the  circle,  and  from  0  a 
secant  is  drawn  meeting  the  circle  in  B  and  the  tangent  in 
C.  The  locus  of  the  projection  of  B  upon  the  ordinate  of  C 
is  called  the  witch. 

To  find  its  equation  take  the  angle  0  =  ^OC  as  a  param- 
eter.    Then  we  have  at  once 

x=  0D  =  AC  =2  a  tsiiie, 

and  y  =  BP  =EB  =  OB  cos  0  =  2  a  cos^  $. 

Eliminating  the  parameter  0,  the  equation  is 

The  form  of  the  equation  shows  that  the  curve  is  symmet- 
rical with  respect  to  the  ?/-axis  and  has  the  a>axis  as  an 
asymptote. 

131.  The  Strophoid.  —  The  distance  of  a  fixed  point  A 
from  a  fixed  line  BC  is  a.  From  A  a  perpendicular  ^0  is 
drawn  to  BC,  also  another  line 
meeting  BC  at  E.  On  this  last 
line  points  P  and  P'  are  taken 
such  that  P'E=EP=OE.  The 
locus  of  the  points  P  and  P'  is 
a  curve  called  the  strophoid. 

To  find  its  locus  take  the  fixed 
line  as  the  2/-axis  and  the  coaxis 
through  the  fixed  point.  Let  the 
coordinates  of  P  and  P'  be  de- 
noted by  (x,  y).  Taking  the  angle 
OAE  as  a  parameter  6,  we  have  at  once  that 

P'E  =  EP=  0E  =  at2ine. 


Hence 
and 


X  =  ±  EP  cos  6  =  ±  a  sin  0  ; 

2/  =  0^  ±  -EPsin  ^  =  a  tan  ^  (1  ±  sin  $). 


164 


ANALYTIC    GEOMETRY 


By  eliminating  the  parameter,  we  have  the  equation, 


y  =±x-—^^, 


Va  — 


X 


or 


y^  =  x^  — —  ■ 
a—  X 


The  curve  is  evidently  symmetrical  with  respect  to  the  a;-axis 
and  has  the  line  x  =  a  as  an  asymptote. 

132.    The  Ovals  of  Cassini.  —  The  locus  of  the  vertex  of  a 
triangle  which  has  a  constant  base  and  has  the  product  of 

its  other  two  sides  equal  to 
a  given  constant  is  called 
the  ovals  of  Cassini. 

Call  the  length  of  the 
base  2  a,  take  the  origin  at 
its  mid-point,  and  let  the 
ic-axis  lie  along  the  base. 
Then  the  conditions  of  the 
problem  make  F'P  -  FP 
equal  to  a  constant,  which  we  call  c^  since  it  is  essentially 
positive. 

Using  the  distance  formula,  we  have 


Y 

___^,«i^ 

(f^^fc^ 

5^^^ 

%^ 

X 

■\/{x  H-  ay  -f  y^  .  -V{x  -  ay  +  ?/"  =  c^- 
Simplifying,  this  reduces  to 

(0)2  +  2/^  +  a^y  _  4  a'^x'^  =  c\ 

Three  types  are  possible,  according  to  the  relative  values 
of  c  and  a.  If  c  =  a,  this  last  equation  may  easily  be  re- 
duced to  the  form, 

{x''^y''y=2a\x^-y% 

which  becomes  on  transformation  to  ^polar  coordinates, 

p2  =  2  a2  cos  2  e, 

showing  that  in  this  case  the  curve  is  a  lemniscafe.     (See 
page  134.) 


HIGHER  PLANE   CURVES 


165 


PROBLEMS 

1.  Transform  into  rectangular  form  the  polar  equations  of  the  fol- 
lowing curves  : 

(a)  the  lima9on  ;  (d)  the  cissoid  ; 

(6)  the  cardioid  ;  (e)   the  four-leafed  rose. 

(c)    the  conchoid ; 

2.  Show  that  the  following  curves  are  not  algebraic  : 

(a)  the  spiral  of  Archimedes  ; 
(&)  the  lituus. 

3.  Plot  the  witch  from  its  parametric  equations. 

4.  Plot  the  strophoid. 

5.  Find  the  polar  equation  of  the  strophoid. 


Ans.  p  =  — 


a  cos  2  d 
cos  6 


6.  Transform  the  rectangular  equation  of  the  ovals  of  Cassini  into 
polar  form . 

7.  Derive  from  the  figure  the  polar  equation  of  the  ovals  of  CassinL 


^1  ^^'(Y-") 


\y^y[y_: 


-jrfr^ 


'  ^^- '       c  "- 


■^ 


r 


^ 


1 


K^'^ 


i!\r 


.-f- 


t  A? 


( 


CHAPTER   X 


TANGENTS  AND  NORMALS 

133.  Definitions.  —  A  line  cutting  a  curve  is  called  a 
secant.  In  the  figure  FQ  is  a  secant.  If  Q  is  made  to 
move  along  the  curve  towards  P,  it  is  clear  that  PQ  will 

turn  about  P  and  approach  a 
limiting  position  PA,  which  is 
called  the  position  of  tangency. 
Therefore 

The  tangent  to  a  curve  at  a 
given  point  is  the  limiting  posi- 
tion of  a  secant  ichen  two  of 
its  intersections  approach  coinci- 
dence at  that  point. 

TJie  normal  to  a  curve  at  a  given  point  is  the  line  perjoen- 
dicular  to  the  tangent  at  that  point. 

134.  Slope  of  Tangent  and  Normal  to  a  Circle  at  a  Given 
Point.  —  To  find  the  slope  of  a  tangent  to  a  curve  at  a  given 
point  Pi(i\,  2/1),  we  first  find 

the  slope  of  the  secant 
through  Pi  and  a  neighbor- 
ing point  P.^(xi  +  h,  2/1  +  k) 
on  the  curve.  This  will  be 
k 
h 

proach  F^  as  a  limit,  the 
secant  approaches  the  tan- 
gent and  the  slope  of  the 
tangent     is     the     limiting 


As  P2  is   made  to   ap- 


TANGENTS  AND  NORMALS  167 

value  of  -,  which  we  write  lim  -,     Since  both  k  and  h  ap- 
h'  h 

proach  zero,  the  limiting  value  of  -  has  the  form  -,  which 

h  0 

is  called  an  indeterminate  form.     To  find  the  value  of  such 

a  limit  special  methods  must  be  employed. 

In  the  case  of  the  circle  we  proceed  as  follows.     Since  P^ 

and  P2  are  both  on  the  circle,  their  coordinates  satisfy  the 

equation  of  the  circle.     Hence 

(x^  +  hy+{yy-\-ky  =  r\ 

Expanding  and  subtracting, 

2  hx^  +  /i2  +  2  ky^  +  k""  =  0, 

h(2x,+  h)=-k(2y,  +  k), 

,  .  k  2  x^  +  h 

and  -  =  —  -— ^ • 

h  2  ?/i  +  k 

From  this  last  relation,  we  have 

lim^  =  -lim?^L±i=_f^. 
h  ^Vi-hk  y^ 

Therefore  for  the  tangent  to  the  circle  x^  -\-  y"^  =  r^, 

m=-^.  (43) 

As  the  normal  is  perpendicular  to  the  tangent,  its  slope 
is  the  negative  reciprocal  of  this  expression,  or  ^  • 

135.  Slope  of  Tangent  and  Normal  to  an  Ellipse  at  a  Given 
Point.  —  The  method  of  procedure  is  the  same  as  that  for 
the  circle.     Let  the  equation  of  the  ellipse  be 


168 


ANALYTIC    GEOMETRY 


and  the  point  of  tangency  be  Pi{Xi,  2/1) •     The  slope  of  a  se- 
cant through  Pi  and  a  neighboring  point  P2{xi-\-h,  2/1  +  k) 

k 


on    the    curve    is   as   before 


h 


Substituting  the  coordinates  of 
Pi  and  P2  in  the  given  equation, 

bW  +  ciV  =  «'&^ 

b\x,  +  hf  +  a\y^  +  ky  =  aW. 
Expanding  and  subtracting, 
2  b^x^h  +  bW  +  2  a^^/iA:  +  a'k^  =  0, 

7i(2  t^xi  +  bVi)  =  -  k(2  ahj^  +  o?k), 


2b%^±¥h 
2  d-yx  +  a?k 


Hence 


,.      k  y      2b%  +  bVi         2b^x^ 

lim  -  =  —  lim  ^— ! =  —  - — -—' 

h  2  ahj,  +  a%          2  ahj^ 


Thus  for  the  tangent  to  the  ellipse,  the  slope  is 


m  = 


a'!/i 


(44) 


The  slope  of  the  normal  is 


136.  Slope  of  Tangent  and  Normal  to  the  Parabola  and 
Hyperbola  at  a  Given  Point.  —  By  applying  the  method  of 
§§  134,  135  to  the  equations  of  the  parabola  and  hyperbola, 
y"^  =  2px  and  6V  _  g^yi  _  ^2^2^  ^g  obtain  the  following 
results : 


Tangent  to  parabola :     m  =  — , 
Tangent  to  hyperbola :    m  =  -^. 


(45) 
(46) 


TANGENTS  AND  NORMALS  169 


The  slopes  of  the  normals  are :  for  the  parabola,  —  ^ ; 


P 


for  the  hyperbola, -^. 

Exercise  1.     Prove  formula  (45). 
Exercise  2.     Prove  formula  (46). 

137.   Equations  of  the  Tangent  and  Normal  at  a  Given  Point. 

—  To  find  the  equation  of  the  tangent  or  normal  to  a  curve  at 
a  given  point,  it  is  necessary  only  to  find  the  slope  and  then 
use  the  point-slope  equation  of  the  straight  line, 

y  -y^  =  m(x-Xi). 

Exercise  3.     Show  that  the  equation  of  the  tangent  to  the  circle 
x2  +  y-  =  r-  at  any  point  Pi(xi,  yi)  reduces  to  XiX  +  yiy  =  f^. 
Hint.  — Since  Pi  is  on  the  circle,  simplify  by  using  x-^-\-yi^=  r^. 

Exercise  4.     Show  that  the  equation  of  the  tangent  at  any  point 
Pi(^i,  y\)  is  : 


(a)  for  the  ellipse      - 
a 

(6)  for  the  hyperbola  - 
a 


'  ,  r  _  1        a;ix  ^ly  _  , 
^  _yl-l        xix_yiy  _  , . 
(c)  for  the  parabola    y-  =  2px,  y\y  =p{x  +  Xi). 


PROBLEMS 

1.   Find  the  slope  of  the  tangent  to  each  of  the  following  curves  foi 
any  point  (a:i,  y\)  on  the  curve  : 

{a)xy  =  '2a^;  cf^^_yl__-i. 

(6)     y  =  ax^;  ^'^  ^   a^     &2  -      -^  ' 

(c)  y^  =  ax^;  (g)  (^x-hy  +  (y-ky  =  r^; 

(d)  x^  =  2py;  (^h)   (y-ky2=2pix-h); 
(0    x^-y'  =  a^;  ^^    (X  -  hy^      (y  -  k^      ^ 

^''         a^       +       b-^       -^' 

Ans.     (a)   -^;  (6)  Sax^^;  (c)     ^^^'"- 


(s)-y;zrr        W  ^^T^^'       W- safety 


170  ANALYTIC    GEOMETRY 

2.  In  each  of  the  following  curves  find  the  slope  of  the  tangent 
and  of  the  normal  at  the  point  indicated  : 

(a)  9  x-'  +  2/2  =  18,  (1,  -  3) ;  (e)  x' +  y'' =  169,  (5,  -  12)  ; 

(&)  x2  +  4  2/2  =  4,  (2,  0);      _  (/)  x'  +  y'' =  25,  (3,  4) ; 

(c)  x2  -  2/2  =  25,     (10,  5y3);  (g)  2y^  =  5  a:,  (10,  5); 

(d)  5x2-12  2/2=13,  (V5,-l);        (h)  ^2  =  36  2/,  (12,  4). 

3.  Write  the  equation  of  the  tangent  and  normal  to  each  of  the 
curves  in  Problem  2  at  the  point  indicated. 

Ans.     (a)  Tangent,  Sx  —  y  —  6  =  0;  normal,  x  +  Sy  +  8  =  0. 

4.  Find  the  points  at  which  the  tangent  to  the  corresponding  curve 
in  Problem  2  has  the  indicated  slope  : 

(«)   -3;         (b)   -iV3;  (c)    -fV3;  (d)   ^. 

(e)   1;  (/)  i;  (g)   - 1 1  W  |. 

Ans.     (a)   (1,  3),  (-  1,-3). 

5.  Find  the  angle  between  each  of  the  following  pairs  of  curves. 
(By  the  angle  between  two  curves  is  meant  the  angle  between  their 
tangents  at  the  points  of  intersection. ) 

(a)  a:2  +  y2  =  25,  (c)   x^  -  y^  =  a\ 

x-\-y   =1  ;  xy  =  V2a^; 

(b)  a:2  +  2/-  =  12,  (d)  ^2  + 4  2/2  =  4, 

2/2  =  X  ;  X  +  2  2/   =  2. 

Ans.     (a)  arc  tan  |. 

6.  Prove  that  the  tangents  at  the  ends  of  the  latus  rectum  of  a 
parabola  are  perpendicular  to  each  other. 

7.  Find  the  angle  between  the  tangents  at  the  ends  of  a  latus 

rectum  of 

(a)  the  ellipse  ;  (6)  the  hyperbola. 

2e 
Ans.     (a)  arc  tan   ^  _    .^. 

8.  Prove  that  the  tangents  at  the  ends  of  any  chord  through  the 
focus  of  a  parabola  are  perpendicular  to  each  other. 

9.  An  ellipse  and  hyperbola  have  the  same  foci.  Prove  that  the 
tangents  at  their  points  of  intersection  are  perpendicular  to  each  other. 

Hint.  — The  value  of  c  is  common  to  the  curves. 

10.  Through  what  point  of  the  ellipse  b-x^  +  a'^y-  =  a^b^  must  a 
tangent  and  a  normal  be  drawn  so  as  to  form  with  the  principal  axis 

an  isosceles  triangle  ?  ,         /      -i  «"  ±  b'^ 

Ans. 


Vrt-i-f  b-y    Va2  4-&2 

11.    Prove  that  a  circle  described  on  a  focal  chord  of  a  parabola  as  a 
diameter  is  tangent  to  the  directrix. 


TANGENTS  AND  NORMALS 


171 


138.    Tangent  and  Normal.     Subtangent  and  Subnormal.  — 
Let  P^The  the  tangent  and  P^iVthe  normal  to  the  curve  in 
the  figure  at  P^.     By  the  length  of  the  tangent  is  meant  the 
distance   from  the  point  of 
tangencj  to  the  point  where 
the  tangent  meets  the  a/'-axis, 
i.e.  the  length  of  P^  T.     Sim- 
ilarly the  length  of  the  nor- 
mal is  the  length  of  P^N. 

The  subtangent  is  the  pro- 
jection of  the  tangent  on  the 
ic-axis,  i.e.  TS.  The  subnor- 
mal is  the  projection  of  the 

normal  on  the  x-axis,  i.e.  JSN.  The  direction  of  reading 
the  subtangent  and  subnormal  is  away  from  the  intersec- 
tion of  the  tangent  with  the  aj-axis  ;  hence  they  are  positive 
if  they  lie  at  the  right  of  this  point ;  negative  if  at  the  left. 

To  find  the  length  of  the  subtangent  and  subnormal,  let 
m  be  the  slope  and  r  the  inclination  of  the  tangent.     Then 

,  SPi       y, 

m  =  tan  t  =  — — ;  =  -^' 

TS      TS 


TS=^ 
m 


Hence 

For  the   subnormal,    the   slope  of   NPi   is 


??i 


-i=tanPi.Y/S'  =  ^=      ^^' 
m  NS 


(47) 


Then 


Therefore  SN  =  my,.  (48) 

To  find  the  lengths  of  the  tangent  and  normal  apply  the 
right  triangle  theorem  to  the  triangles  NP^S  and  SPiT. 

139.  Tangent  having  a  Given  Slope.  —  Suppose  it  is  re- 
quired to  find  the  tangent  to  the  curve  a;^  +  4  ?/2  =  8  having 
the  slope  ^. 


172 


ANALYTIC   GEOMETRY 


Here  tlie  slope  is  given  and  it  is  necessary  to  find  the 
point  of  contact,  which  vre  call  Pi{x^,  y^).  Since  P^  is  on 
the  curve, 


a^i'  +  4?/i2  =  8. 


(«) 


Now  a;2  +  4  2/-  =  8  is  an   ellipse,   for   which   a^  =  8    and 
¥  =  2.     Hence  by  (44)  the  slope  of   the  tangent  at  F^  is 

-^     But  by  the    con- 

ditions  of  the  problem  the 
slope  is  to  be  i.     Hence 

Sy~2' 


iP) 


Solving  (a)  and  (6) 
simultaneoush',  we  get 
X,  =  t2,  2/x  =  ±  1.     Thus 

there  are  two  tangents  with  points  of  contact  (2,  —  1)  and 

(—  2,  1)  and  equations 

y±l  =  \{xT  2), 
which  reduce  to 

.T  -  2  ?/  T  4  =  0. 

The  method  illustrated  in  the  above  problem  is  typical 
and  may  be  applied  to  any  problem  of  the  same  character. 

140.  Tangent  from  a  Given  External  Point.  —  To  find  the 
equation  of  a  tangent  to  a  curve  from  a  given  external  point 
we  proceed  as  in  the  following  problem. 

Let  the  curve  be  ic^  _|_  4  ^2  _  g  g^j-^j^^  ]^\^q  point  (1,  f ).  As 
before,  let  the  point  of  contact  be  P^ix^,  y^.     Then  we  have 

a;,2  +  4  2/i^=8.  (a) 

2  0^1 


The  slope  of  the  tangent  is,  by  (44),  - 
equation  of  the  tangent  is 

Zi  X-i  /  \ 

y-yi  =  -^{^-^i)' 
82/i 


82/1 


Hence  the 


(p) 


TANGENTS  AND  NORMALS  173 

Since  tliis  is  to  pass  through  (1,  |),  we  have 

82/1 

Solving  equations  (a)  and  (b)  simultaneously,  we  get 

a^i  =  2  or  -  I,      yi  =  loT  I. 

Thus  there  are  two  tan- 
gents with  points  of  contact 
(2,1),  and  (-1,1).  For  the 
first  of  these  the  slope  is 
—  4-  and  the  equation 

2/-l=-^(x-2), 

or  x-\-2y— 4^  =  0. 

Eor  the  second  the  slope  is  Jj  and  the  equation 

or  a;  -  14  ^  +  20  =  0. 


PROBLEMS 

1.  Write  the  equations  of  the  tangent  and  normal  to  each  of  the 
following  curves  at  the  point  indicated  : 

(a)  x^  +  y^  =  lS,  (2,  -3);  (e)  y  =  2x^,  (2,  16); 

(6)  x2  +  2y2=18,  (4,  -1);  (/)  y^=-2x^  (-2,  -4); 

(c)  x2  -  2y2  =  18,  (_  6,  3)  ;  (g)  V  =  9x^  (1,  9)  ; 

(d)  2  2/2  +  5  X  =  0,  (-  10,  5)  ;  (h)  xy  =  24,  (4,  6). 

2.  Find   the  length  of   the  subtangent   and   subnormal  for  each 
curve  in  Problem  1. 

3.  Find  the  equation  of  the  tangent  to  each  of  the  curves  in  Prob- 
lem 1  parallel  to  the  line  4  .r  —  3  ?/  =  6. 

4.  Find  the  equation  of  the  tangent  from  the  point  (5,  —  1)  to 
each  of  the  curves  (a),  (6),  (c),  and  (d)  in  Problem  1. 


174  ANALYTIC   GEOMETRY 

6.  Find  the  lengths  of  the  subtangent,  subnormal,  tangent,  and 
normal  at  any  point  (x,  y)  of 

(a)  2/2  =  2px  ;  (c)  b'^x^  -  aY~  =  «^&^  5 

(6)  b-^x'^  +  a^y^  =  a'^b^ ;  (d)  x^  +  y^  =  r^. 

Ans.   Subtangent:   (a)  2x,  (b)  ~^,   (c)   ^,  (d)   -^';  sub- 

b-x  b'-x  X 

normal:  (a)  p,  (b)   --^,   (c)  "Y'   ^^^   ~  ^• 

6.  Write  and  simplify  the  equation  of  the  normal  to  each  of  the 
curves  of  Problem  5  at  any  point  (xi,  yi). 

7.  Show  that  a  line  from  the  focus  of  the  parabola  y^  =  2px  to 
the  point  where  any  tangent  cuts  the  ?/-axis  is  perpendicular  to  the 
tangent. 

8.  Prove  that  the  tangents  at  the  ends  of  a  latus  rectum  intersect 
on  the  directrix  in  the  case  of  : 

(a)  the  parabola  ;    (&)  the  ellipse  ;    (c)  the  hyperbola. 

9.  Prove  that  the  tangents  to  the  ellipse  6-x'-  +  cfiy"^  =  a^ft'-^  and  the 
circle  r?  +  y^  =  a^  at  points  having  the  same  abscissa  meet  on  the 
cc-axis. 

10.  Prove  that  the  triangle  formed  by  a  tangent  to  the  curve 
2  xy  =  a^,  the  a:-axis,  and  a  line  joining  the  point  of  contact  with  the 
origin  is  isosceles. 

11.  Prove  that  the  triangle  formed  by  the  tangent  to  the  curve 
2  xy  =  a'^  at  any  point  and  the  coordinate  axes  has  the  area  a^. 

12.  A  tangent  to  the  parabola  y-  =  2px  has  the  intercepts  on  the 
axes  numerically  equal.     Find  its  equation. 

13.  If  a  tangent  to  a  parabola  meets  the  latus  rectum  produced  at 
A  and  the  directrix  at  B,  prove  that  FA  =  FB,  F  being  the  focus. 

14.  At  what  points  on  the  ellipse  ^  +  ^  =  1  are  the  tangents 
equally  inclined  to  both  axes  ?  ^ 

15.  If  d  and  d'  are  the  distances  of  the  foci  of  an  ellipse  from  a 
tangent,  prove  that  dd'  is  the  square  of  the  semi-minor  axes. 


141.    The  Parabolic  Reflector.  —  Let  TP  be  tangent  to  the 
parabola  i/  =  2  px  at  Pix,  y),  and  let  TM  be  the  subtangent 


TANGENTS  AND  NORMALS 


175 


Applying  tlie  subtangent  formula,  TM  =  2  x;  hence,  TO 
=  03/  =  X.     Xow  if  F  is  the  focus,  this  gives 

TF=x+P.  ^^' 

But      FP=  XP=x+^^. 


Hence,  TPF  is  isosceles, 
i.e.  the  tangent  to  a  ■parabola 
makes  equal  angles  icith  the 
focal  radius  and  the  prin- 
cipal axis. 

This  i3rinciple  is  used  in  the  construction  of  parabolic 
reflectors.  These  are  reflectors  whose  surfaces  are  gen- 
erated by  revolving  a  parabola  about  its  principal  axis. 
The  lamp  is  placed  at  the  focus.  Then  by  the  laws  of  optics 
a  ray  of  light  from  F  to  any  point  on  the  surface  P  is  re- 
flected along  PE  so  that  angle  TPi^=  angle  HPE.  But 
this  makes  angle  HPE  equal  to  angle  PTF  by  the  above 
proof.  Therefore  PE  is  parallel  to  OX.  Thus  all  the  rays 
of  light  from  a  lamp  placed  at  F  are  reflected  along  lines 
parallel  to  the  axis. 

Exercise  5.  V&e  the  above  property  of  the  tangent  to  devise  a 
method  of  constructing  a  tangent  to  a  given  parabola  at  any  point  by 
ruler  and  compasses. 


142.   Angle  between  the  Focal  Radii  and  the  Tangent  to  an 
Ellipse  at  any  Point.  —  Let  PT  be  a  tangent  and  PX  sl 

normal  to  the  ellipse  at 
any  point  P(x,  y)  on  the 
ellipse,  whose  foci  are  F 
and  F\  We  desire  to  show 
that  these  lines  make  equal 
angles  with  the  focal  radii 
to  P. 


176  ANALYTIC    GEOMETRY 

The  slope  of  FP  is  -1—,  of  FP,      ■'    ,  and  of  NP.  ^^, 

X'  +  c  X  —  c  '  b-x 

by  §  135.     Hence 

a^y         y 


tan  F'P\=    ^'^      ^  +  ^     =  g^x?/  +  g^cy  -  b'xy 
^  ay  62^2  +  b'^cx  +  a2?/2 

62a;(x  +  c) 

~  a^b^  -f  62cx  ~  62 ' 


and  tan  FPN  = 


y       «^.y 

X  —   C           62^              62^?/  — 

a2a;j^  _p  ci2c2/ 

62X(X  —  C) 

62cx  +  a22,2 

a2c?/  —  c2x?/      cy 

aW  -  b'^cx      62- 

Therefore  FPN=  FPN\   and  the  angles   made   with  the 
tangent,  being  complementary  to  these,  are  also  equal. 

Exercise  6.     Show  how  to  draw  a  tangent  to  an  ellipse  at  any  point. 

143.  Diameters  of  a  Conic.  —  The  locus  of  the  middle 
points  of  a  system  of  parallel  chords  of  any  conic  is  called 
a  diameter  of  the  conic. 

Let  a  chord  with  slope  tn  meet  the  ellipse 

^'  4-  2/'  =  1 
a2      62 


TANGENTS  AND  NORMALS  177 

in  the    points    Pi(x^,  2/1)  and  P^^x^^  +  h,  y^  +  A-).     Then  by 
§  135  the  slope  of  the  chord  is 


a\2y,  +  k) 


Now   let   P(x,   y)    be    the 
mid-point  of  the  chord  ;  then 


2  .X  =  2  X,  H-  h 
and        2y  =  2y,  +  lc 

by  the  mid-point  formulas. 

Substituting,  we  have,        m  =  — 


P„{x+h,xj^-^k) 


b'^x 
a-y 


or 


(49) 


which  is    the  equation  of   the  locus.     From  this   equation 
we  see  that  the  diameter  is  a  straight  line  through  the  cen- 

ter  of  the  ellipse  of  slope  m'  = .     Transferring  the  m 


a^m 


to  the  other  side,  we  have  an  important  relation  between 
the  slopes  of  a  diameter  and  its  chords, 

&2 


mm  =  — 


(SO) 


In  similar  manner,  we  may  show  that  the  equation  of  the 
diameter  of  the  parabola  2/^  =  2  pa;,  is 


m 


(51) 


where  m  is  the  slope  of  the  system  of  chords. 

For  the  hyperbola  we  obtain  as  the  equation  of  the  diameter 
of  a  set  of  chords  of  slope  m, 

&2 


y-^ 


a^m 


X. 


(52) 


178 


ANALYTIC    GEOMETRY 


The  relation  between  the  slopes  of  the  chords  and  their 
diameter  is 

(53) 


mm  =  —  • 


Exercise  7.     Prove  (51). 
Exercise  8.     Prove  (52). 

144.   Conjugate  Diameters  of  an  Ellipse.  —  The  relation 

mm  = 

o? 

states  the  condition  satisfied  by  m  and  m'  if  the  line 
y  =  m'x  is  the  diameter  of  a  set  of  chords  of  slope  m.  But 
as  m  and  m'  can  be  interchanged  in  (50)  without  altering 
the  relation,  we  see  that  y  =  mx  is  the  diameter  of  the  set 
of  chords  of  slope  m\  parallel  to  the  first  diameter. 

Two  diameters  of  slopes  m  and  ??i',  such  that  mm^  = 


are  called  conjugate  diameters. 


by  §  135.     But  ^  = 
2/1 


a' 

Each  diameter  is  one  of  the 
chords  bisected  by  its  con- 
jugate. 

It  is  readily  seen  that 
tangents  at  the  ends  of  a 
diameter  are  parallel  to  the 
conjugate  diameter. 

For,  let  PifcTi,  ?/i)  be  the  end 
of  the  diameter  y  =  mx.  The 
slope  of  the  tangent  at  Pi  is 

-,    since    Pi   lies   on   y  =  mx. 


62 


cr/>i 


,  which  is  also  the 


Hence  the  slope  of  the  tangent  is 
slope  of  the  conjugate  diameter. 

This  property,  together  with  Exercise  6,  supplies  a  method 
of  constructing  a  diameter  conjugate  to  a  given  diameter 
with  ruler  and  compasses. 


TANGENTS  AND  NORMALS 


179 


145.  Conjugate  Diameters  of  the  Hyperbola.  —  The  relation 
between  the  slope  of  a  diameter  of  the  hyperbola  W-:i? 
—  ahp-  =  a262  and  the  slope  of  the  corresponding  chords  has 
been  found  to  be 


mm  =—• 
a2 


As  the  equation  of  the  conjugate  hyperbola,  —  b^x"^  +  aV 
=  aW,  is  obtained  by  substituting  —  5^  for  b^  and  cv^  for 
—  a2,  it  is  easy  to  see  that  the  corresponding  relation  for 
the   diameter  and    chords    of   the    conjugate   hyperbola   is 

-  =  — .     That  is,  it  is  the  same  in  both  the  hy- 


mni'  = 


perbola  and  its  conjugate.     As  in  the  case  of  the  ellipse,  two 

diameters  of  slopes  m  and  m'  such  that  mm'  =  —  are  called 
conjugate  diameters. 


Similarly,  the  chords  bisected  by  the  diameter  y  =  mx  are 
parallel  to  the  conjugate  diameter  y  =  vi'x.     Also  the  tan- 


180  ANALYTIC   GEOMETRY 

gents  to  a  hyperbola  at  the  ends  of  a  diameter  are  parallel  to 
the  conjugate  diameter. 

Exercise  9.     Prove  that  the  tangents  at  the  ends  of  a  diameter  of  a 
hyperbola  are  parallel  to  the  conjugate  diameter. 

PROBLEMS 

1.  Find  the  diameter  bisecting  the  system  of  chords  of  slope  2  in 
each  of  the  following  conies : 

(a)  a;2  +  4  1/2  =  16  ;  (d)  2y^  -  6x  =  0; 

(6)  x^  -92/2  =  9;  (e)  9 x^  -\-Wy^  =  144  ; 

(c)  a:2  +  y^  =  18  ;  (/)   16  x^  -9y^=  144. 

2.  Find  the  diameters  conjugate  to  those  obtained  in  Problem  1. 

3.  What  is  the  relation  holding  between  conjugate  diameters  of 
the  ellipse  U^y^  +  a^x-  =  a-b'^  ? 

4.  Show  that  conjugate  diameters  of  an  ellipse  lie  in  different 
quadrants. 

5.  Show  that  conjugate  diameters  of  a  hyperbola  lie  in  the  same 
quadrant. 

6.  Prove  that  the  major  axis  of  an  ellipse  is  greater  than  any 
other  diameter. 

Hint.  —  Show  that  4  a^  is  greater  than  the  square  of  any  diameter. 

7.  Lines  are  drawn  joining  the  ends  of  the  major  and  minor  axes 
of  an  ellipse.     Show  that  the  diameters  parallel  to  these  are  conjugate. 

8.  Prove  that  the  tangents  to  an  ellipse  at  the  ends  of  a  diameter 
make  equal  angles  with  lines  joining  these  points  to  a  focus. 

9.  Prove  that  the  tangents  to  two  conjugate  hyperbolas  at  the  ends 
of  a  pair  of  conjugate  diameters  meet  on  an  asymptote. 

10.  Prove  that  if  any  pair  of  conjugate  diameters  of  a  hyperbola 
are  equal,  the  hyperbola  is  equilateral. 

11.  I'rove  that  in  an  equilateral  hyperbola  the  asymptotes  bisect 
the  angle  between  any  pair  of  conjugate  diameters. 


^ 


CHAPTER   XI 

SOLID   ANALYTIC  GEOMETRY 

146.  Plane  Analytic  Geometry  deals  with  the  properties 
of  figures  which  are  wholly  contained  within  a  single  plane. 
In  Solid  Analytic  Geometry  this  restriction  is  removed. 
The  effect  of  the  addition  of  one  dimension  is  to  generalize 
the  work  of  the  previous  chapters,  the  number  of  variables 
being  increased  to  three,  the  loci  of  equations  being  surfaces, 
etc.  The  student  should  note  carefully  the  correspondences 
between  the  two  systems. 


147.  Coordinates.  —  The  position  of  a  point  is  determined 
with  reference  to  three  mutually  perpendicular  planes,  XO  F, 
YOZ,  and  XOZ.  These  are 
called  the  coordinate  jolanes, 
and  designated  as  the  xy-, 
yz-,  and  xz-planes  respec- 
tively. Their  lines  of  inter- 
section, OX,  OY,  and  OZ, 
are  called  coordinate  axes  and 
designated  as  the  x-,  y-,  and 
z-axes  respectively ;  and  their 
common  point,  0,  is  called 
the  origin. 

The  coordinates  of  a  point  P  are  its  distances  from  the  re- 
spective coordinate  planes  measured  parallel  to  the  coordinate 
axes.  The  x-coordinate  is  the  distance  from  the  ^/^-plane 
parallel  to  the  a;-axis,  the  y-coordinate  that  from  the  a'Z-plane 
parallel  to  the  2/-axis,  and  the  z-coordinate  that  from  the 

181 


Z 

+ 

c 

D 

/ 

/ 

-t 

p 

+  , 

__ 

4.        X^ 

B 

/ 

• 
/ 
/ 

/ 

E 


182 


ANALYTIC    GEOMETRY 


icy-plane   parallel  to  the  2;-axis.      For   the  point  P  in  the 
figure, 

x  =  FP=BE=CD=  OA, 

y  =  DP  =  AE  =  CF  =  OB, 

z  =  EP  =  BF  =  AD  =  OC. 

Corresponding  to  each  point  there  are  evidently  always 
three  coordinates.     Conversely,  a  point  is  completely  located 

by  its  coordinates.  For  the 
ic-coordinate  fixes  it  in  a 
plane  parallel  to  the  yz-iplsme 
(or  perpendicular  to  the  x- 
axis),  and  similarly  for  the 
other  coordinates  ;  and  these 
three  planes  meet  in  but  one 
point.  In  plotting  a  point, 
we  usually  take 

^+  x=  OA,     y  =  AE,     z  =  EP. 


Z 

■+ 

/ 

c 

D 

/ 

// 

F 

/ 
/ 

• 

/ 

/ 
/ 
/ 
/ 
/ 
/ 

P 

+  . 

B. 

i      x' 

E 


The  positive  directions  are  as  indicated  in  the  figure. 
The  coordinate  planes  divide  space  into  eight  octants,  which 
are  distinguished  by  the  signs  of  the  coordinates  of  the 
points  within  them.  For  example  the  octant  containing  P 
in  the  figure  has  all  the  coordinates  positive  in  sign. 

148.  Radius  Vector  and  Direction  Cosines.  —  TJie  distance 
of  a  point  from  the  origin  is  called  its  radius  vector,  and  is  de- 
noted by  p. 

For  the  point  P  in  the 
figure  p  =  OP.  It  is  evi- 
dent that 

p'-^OW  +  eP 

=  02'  +  'aW  +  EP". 

Hence 

p2  =  x2  +  y^  +  z\      (54) 


SOLID   ANALYTIC   GEOMETRY 


183 


The  angles  between  the  line  OP  and  the  positive  halves 
of  the  X-,  y-  and  2;-axes  are  called  the  direction  angles  of  the 
line  OP  and  are  denoted  by  a,  ^,  and  y,  respectively.  The 
cosines  of  these  angles  are  called  the  direction  cosines  of 
the  line.  They  evidently  fix  its  direction.  From  the  figure 
it  is  evident  that 

X  =  p  cos  a,  1/  =  p  cos  p,  z  =  p  cos  7.  (55) 

Formulas  (54)  and  (55)  give  at  once  the  important  relation 

cos^  a  4-  COS"  p  -h  cos-  y  =  I.  (56) 

It  is  sometimes  convenient  to  locate  a  point  by  means  of 
its  radius  vector  and  its  direction  angles.     In  this  case  we 

P(x,  y,  z)  =  P{p,  a,  13,  y). 

This  mode  of  representation  is  analogous  to  the  polar  coor- 
dinate system  in  plane  analytic  geometry. 


149.  Distance  between  Two  Points.  —  The  distance  between 
any  two  points  Pi(xi,  y^,  z{)  and  P2(o^2,  2/2?  ^2)  ^^  given  by  the 
formula, 


d  =  V(x,  -  x^y  +  {y,  -  yO'  +  (2i  -  22)-. 

z 


To  prove  this,  pass 
planes  through  the  given 
points  Pi  and  P^,  j)arallel 
to  the  coordinate  planes. 
These  will  form  a  rec- 
tangular parallelopiped,  of  } 
which  P1P2  is  the  diagonal. 

By  elementary  geometry, 


P1P2  =  P.A  -f-  AW  +  DP(. 

But  P^A  =  iCi  —  x^,  AD  =  yx  —  y-i,  and  DPj  =  Zi  —  z^. 
Hence  we  have  the  formula. 


(57) 


184  ANALYTIC   GEOMETRY 

150.  Direction  of  a  Line.  —  The  direction  angles  of  a  line 
not  passing  through  the  origin  are  defined  as  the  direction  angles 
of  a  jKirallel  line  through  the  origin  icith  the  same  positive 
direction. 

The  positive  direction  on  a  line  is  arbitrary  and  indicated 
by  the  order  in  which  its  end-points  are  read.  It  is  evident 
that  if  the  direction  is  changed,  the  direction  angles  are  re- 
placed by  their  supplements.  Thus,  if  the  direction  angles 
of  PoA  are  a,  /?,  and  y,  those  of  P^Pn  are  tt  —  a,  tt  —  ^,  and 
TT  —  y.  In  all  cases  the  direction  angles  are  in  value  between 
0  and  TT. 

In  the  above  figure  the  edges  of  the  parallelepiped  are 
parallel  to  the  coordinate  axes ;  hence  the  direction  angles 
of  the  line  PoPi  are : 

a  =  AP^P^,  /?  =  BP^P,,  y  =  CP.Pi- 

Then  we  have  at  once 

cos  a  =    ^    ,    -,  cos  6  =  ^^— ^- ,  cos  v  =       ,     •      (^") 
a  ^  d  'a 

Since  the  direction  cosines  are  connected  by  formula  (56) , 
they  may  be  found  if  three  numbers  to  which  they  are  pro- 
portional are  known.     For  if 

cos  a  :  cos  ^  :  cos  y  =  a  :  b  :  c, 
then 

cos-  a  _  cos^  ^  _  cos'^y  _  cos-  a  +  cos'^  ^  -f-  cos^  y  _  1 

a"    ~     U"     ~     &     "  a'  +  62_|_c2  ~  a'' ^h^ -\-c^' 

whence 

a  ^ 

cos  a  = ,  cos  p  = 


Va2  4-62_|_c2  V«2  +  62  _^  c* 

c 

cos  y 


Va2  +  62+c2 

Any  three  numbers  proportional  to  the  direction  cosines  of 
a  line  are  called  direction  numbers   of  the  line. 


/ 

SOLID   ANALYTIC  GEOMETRY 


185 


151.  Angle  between  Two  Lines.  — -  Tlie  angle  between  two 
lines  which  do  not  meet  is  defined  as  the  angle  between  two 
intersecting  lines  parallel  to  the  given  lines  and  having 
the  same  positive  directions.  If  the  lines  are  parallel,  the 
angle  between  them  is  0  or  tt, 
according  to  their  directions. 

We  now  derive  a  formula  for 
the  angle  between  two  lines  in 
terms  of  their  direction  cosines. 
Let  the  lines  through  the  origin 
parallel  to  the  given  lines  be 
OPi  and  OP2,  where  the  coordi- 
nates of  Pi  and  P2  ^1*6  (Xi,  y^,  z^) 
and  (x2, 2/2?  ^2)  respectively.  Let 
d  be  the  distance  between  these  points ;  let  pi  and  pj  ^6 
their  radius  vectors  ;  and  let  6  be  the  angle  between  the 

lines.     Then                             2  _i_  ^  2      ^2 
cos  6  =  ^^    '  ^■' 

2  pxp2 


z 

'^ 

7 

/ 

\ 

0 

A 

■^ 

i<^  ' 

But 


Pi'  =  ^1'  +  2/1'  +  ^x\ 
P2^=^X2^  +  y.^-\-Z2\ 

^2 = {x,  -  x^y + (2/1  -  2/2)' + (2!i  -  22)'. 

.-.  cos  ^=:Ei^2_+ML±iii2. 
P1P2 

Now  from  (55)  —  =  cos  «!,  etc.     Hence  this  last  equation 
becomes  ^^ 

cos  9  =  cos  ai  cos  a2  +  cos  pi  cos  P2  +  cos  71  cos  72-    (5^) 
When  the  two  lines  are  parallel,  we  have 


DT 


Wl  =  TT- 


Cto 


ft=7r-^2, 


and  yi  =  72, 

and  yi  =  TT  —  y2. 


When  the  two  lines  are  perpendicular,  0  =  '^  and  (59)  becomes 

Li 

cos  tti  cos  a2  +  cos  pi  cos  p2  4-  cos  ^i  cos  72  =  0.   (60) 


^3 


^•^-^ 


b 


Wr-  ■'. 

186  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Plot  the  points  : 
{a)   (3,4,5),  (6,0,0),  (-6,1,3); 

j^)   (-2,3,-1),  (7,-4,5),  (1,0,  1); 
-^c)    (0,0,2),  (0,3,-4),  (-2,  -2,  1); 

2.  For  each  of  the  above  points  find  the  radius  vector  and  its 
direction  cosines. 

3.  Generalize  the  definition  of  symmetry  with  respect  to  a  line  so 
that  it  will  apply  to  symmetry  with  respect  to  a  plane.  Show  that  the 
point  (a*,  y,  s)  is  symmetrical  to  {—x^y^z)  with  respect  to  the  ?/2;-plane  ; 
to  {x^  —  ?/,  2)  with  respect  to  the  cc2-plane ;  to  (a:,  ?/,  —  z)  with  respect 
to  the  a:2/-plane. 

4.  Show  that  the  point  {x,  ?/,  z)  is  symmetrical  to  (—  x,  —  ?/,  z) 
with  respect  to  the  ^-axis  ;  to  (—  a;,  y,  —  z)  with  respect  to  the  ?/-axis  ; 
to  (.r,  —  y^—  z)  with   respect  to  the  x-axis. 

5.  Show  that  the  following  points  are  vertices  of  a  regular 
tetrahedron :         -  ^^  ^  >.. 

(a)   (4,0,0),  (0,  4,'0),  (0,^4),  (4,^4,4); 

(6)   (6,  -  2,  1),  (3,  1,  1),  (3,  -  2,  4),  (2,  -  3,  0). 

6.  Show  that  the  three  points  (1,  4,  -  2),  (4,  7,  1),  (—  2,  1,  -  5) 
are  m  a  straight  line.  '     ^^  ^  ^  ^-^ 

iW  Show  by  tw^o  methods  that  the  following  are  the  vertices  of  a 
rignt  triangle : 

(a)  (-  2,3,5),  (-5,-3,3),  (0,0,  11); 

(b)  (2,6,-5),  (-1,0,-7),  (4,3,  1). 

//^    Find  the  angles  of  the  triangles  whose  vertices  are  given  in 
Pr^Jolem  1. 

9.    Find  the  coordinates  of  the  point  determined  as  follows : 

/)  =  8,    C0Sa=:f,    C0S7=—  A. 

10.  Show  that  the  coordinates  of  the  point  Pq  dividing  the  line 
P1P2  in  the  ratio  n  :  r-z  are 

_  r2Xi  +  riXo         _  T'^yi  4-  ri?/2         _  ^2^1  +  r\Z2 
'''-     n  +  r,    '  y'-     ri  +  r2    '   '' '      n  +  r^   ' 

11.  What  is  the  locus  of  points  for  which 

(a)  2  =  0;  (e)  x  =  3,  2=-  2; 

\b)  y  =  &  constant ;  (/)  p  =  6; 

(c)  y  z=z  x  =  0  ;  ■  (g)  cos  a  =  0  ; 

(d)  xz=y;  '  (h)  cosa  =  cos/3  =  0  ? 


SOLID  ANALYTIC   GEOMETRY  187 


12.    "What  is  the  projection  of  P(  -  2,  3,  —  4)  on 

(a)  the  icy-plane  ;  >-^  (^ 

(6)  the  0-axis  ? 


>-^r 


152.  The  Locus  in  Solid  Geometry.  —  Tlie  locus  of  a  point 
in  space  satisfying  a  given  set  of  conditions  is  in  general  a 
surface  or  group  of  surfaces.  To  illustrate,  consider  the  locus 
of  a  point  at  a  given  distance  a  from  a  given  fixed  point  C. 
This  is  evidently  the  definition  of  the  surface  of  a  sphere  of 
radius  a  and  center  C.  Again  the  locus  of  a  point  at  a 
constant  distance  a  from  a  given  straight  line  I  is  evidently 
a  circular  cylindrical  surface  having  I  as  an  axis  and  a  as  a 
radius. 

Extending  the  definition  of  the  equation  of  a  locus  given  in 
Chapter  I,  this  shows  that  the  locus  of  a  single  equation  in 
the  three  variables,  x,  y,  and  z,  is  in  general  a  surface. 

If  the  coordinates  of  a  point  satisfy  tico  equations 
simultaneously,  its  locus  must  be  common  to  the  two 
surfaces  which  are  the  loci  of  the  equations,  i.e.  it  must 
be  their  intersection,  which  is  a  curve  or  set  of  curves. 
Finally  the  points  determined  by  a  set  of  three  equations 
will  be  the  intersections  of  the  surface  determined  by 
the  third  with  the  curves  determined  by  the  first  two, 
and  hence  will  be  isolated  points.  Note  the  analogy  to 
plane  analytic  geometry,  where  one  equation  determines  a 
curve,  and  tico  equations  the  isolated  points  in  which  the 
curves  meet. 

153.  The  Normal  Equation  of  the  Plane.  —  The  perpen- 
dicular upon  a  plane  from  the  origin  is  known  as  the 
normal  axis  and  the  distance  of  the  plane  from  the  origin 
is  known  as  the  normal  inter cej^t,  A  plane  is  completely 
determined  if  the  length  of  the  normal  intercept  p,  and 
the  direction  angles,  «,  /3,  and  y,  of  the  normal  axis  are 
known. 


188 


ANALYTIC   GEOMETRY 


To  find  the  normal  equa- 
tion of  a  plane,  take  any 
point  P{x,  y,  z)  in  the 
plane  and  join  it  to  the 
origin.  Draw  the  normal 
axis  ON.  Let  p^  be  the 
radius  vector  of  P,  and  «i, 
ySi,.  yi,  its  direction  angles  ; 
let  «,  /8,  y  be  the  direc- 
tion angles  of  ON  \  and 
let  B  be  the  angle  between  OP  and  ON. 
Then  by  (59), 

cos  Q  =  cos  «!  cos  a  +  cos  /S^  cos  /5  -f  cos  y,  cos  y. 

But  GOS0  =  —=^- 

OP      Pi 

Eliminating  6  between  these  equations, 

Pi[cos  «i  cos  a  +  cos  /5i  cos  /3  +  cos  yi  cos  y]  =  p. 

This  becomes  on  applying  (55), 

X  cos  a  +  y  cos  p  +  z  cos  y  =  p,  (61) 

which  is  the  normal  equation.  ^     ^ 

Exercise.  — Show  that  for  any  point  P(x,  ?/,  z)  not  in  the  plane 
X  cos  a  -\-  y  cos  /3  +  <:  cos  y^  p. 


154.   Plane  Parallel  to  One  or  More  Coordinate  Axes.  —  Sup- 
pose the  plane  is  given  parallel  to  the  2;-axis.     In  this  case 

the  normal  axis  will  lie  in  the  .ry-plane  and  y  =  -•     Hence 
(61)  reduces  to  the  form  V--  r 


X  cos  a  +  !/  cos  p  =  />. 


(61a) 


Similar  equations  are  obtained  for  planes  parallel  to  either 
of  the  other  axes. 


^  ^    ,  d--  6/) 

^     SOLID   ANALYTIC   GEOMETRY  189' 


In  case  the  plane  is  parallel  to  both  the  y-  and  2-axes,  we 

TT 

2 


have  the  normal  along   the  a;-axis.      Then  y8=y  =  |,  and 


a  =  0.     For  this  case  (61)  reduces  to  the  form 

x=p, 
a  result  obvious  from  the  definition  of  coordinates. 

155.   The  General  Equation  of  the  First  Degree.  —  The  nor- 
mal equation  (61)  of  the  plane  is  of  the  first  degree.    We  now 
wish  to  prove  conversely  that  the  locus  of  the  general  first-  J^ 
degree  equation      Ax  +  By -\- Cz  +  D  =  (i  (62)  ^5 

is  a  plane.     To  do  this  we  simply  show  that  (62)  can  al- 
ways be  reduced  to  the  form  (61). 

Dividing  both  sides  of  the  equation  by  ±  V^^  ■{-  B^-\-  C^^ 
we  have  J^  d^^. 

^■■-         :.  +  -  ^  -.■  ' 


-D 


By  §  150  the  coefficients  of  .^',  y,  and  z  a're^irection  cosines 
of  a  line.  Hence  this  equation  has  the  same  form  as  (61) 
and  its  locus  is  a  plane.     The  direction  cosines  of  its  normal 

are 

A  B  ^  C 

±  V^42  ^B'+  C'  '  ±  V^l^+5-+C2  '  _t  ^j2  +  ^^  +  (72 
and  the  length  of  its  normal  intercept  is 

-D 

±  -Va"  +  b'+c^' 

The  sign  of  the  radical  is  taken  opposite  to  that  of  D  so 
that  the  normal  distance  p  shall  be  positive.  Comparison 
with  (61a)  shows  that  if  any  one  of  the  three  variables  is 
missing  the  locus  is  a  plane  parallel  to  the  corresponding  axis. 


190  ANALYTIC   GEOMETRY 

156.  Angle  between  Two  Planes.  —  The  angle  between 
two  planes  is  readily  seen  to  be  the  same  as  that  between 
their  normal  axes.  Hence  the  angle  between  two  planes 
can  be  found  by  formula  (o9).  If  we  wish  to  express  cos  d 
in  terms  of  the  coefficients  of  (62),  substitution  of  the  above 
values  of  the  direction  cosines  gives 

AA'  +  BB'  -I-  Ca  ,no. 

cos  0  =  —  -— .  (63) 

V^-  +  52  +  C2 .  \A''  +  5'2  +  C 

Parallel  Planes.  —  If  two  planes  are  parallel,  their  normal 
axes  are  the  same.  Hence  their  direction  cosines  are  the 
same  or  numerically  equal  with  unlike  signs.     This  gives 

A  ±A'  B  . 

=  etc. 


By  alternation  we  have 


A 

V^2  +  52  ^  (72 
VA'^  +  B'^+  C""' 

B  _  V^42H-52  +  C2 

A' 

B'          V^'2_|_^/2_^(7/2 

C              V^2  +  52  _^   (12 

C  V^'2   _^   ^/2   ^  (^/2 

This  gives  as  the  condition  for  parallelism, 

^  =  ^  =  ^ 

A'      B'      C 


(64) 


Perpendicular  Planes.  —  If  the  planes  are  perpendicular, 
cos  ^  =  0.     Then  from  (63)  we  have  at  once 

AA'  +  BB'  +  CC  =  0.  (60) 

157.  The  Intercept  Equation.  —  The  intercepts  of  a  plane 
are  the  distances  from  the  origin  to  the  points  in  which  it 
meets  the  x-,  y-,  and  2;-axes.  They  are  denoted  by  a,  b, 
and  c,  respectively. 

If  in  equation  (62)  we  set  y  =  z  =  0,  we  find   that  the 


\ 

SOLID  ANALYTIC   GEOMETRY  191 

i»-intercept,  a,  is .     Similarly  6  =  —  —,  and   c  = 

^  A  ^  B  C 

By  transposing  D  and   dividing   both  sides  by  —  D,   (62) 

may  be  reduced  at  once  to  the  form 

-  +  l  +  ~  =  l,  (66) 

a      b     c 

which  is  knoT^Ti  as  the  intercept  form  of  the  equation. 


PROBLEMS 

1.  Write  the  equations  of  the  following  planes  in  the  intercept  and 
normal  forms  and  determine  which  octant  contains  the  foot  of  the 
normal  axis. 

(a)3a:-4?/+82;-5  =  0;  (d)4x+5?/-20  =  0; 

(&)4x+?/  +  2-8  =  0;  (e)x  +  s-l  =  0; 

(c)  3  X  -  y  -  8  0  +  10  =  0  ;  (/)  X  +  ?/  +  2  =  12.  ^ " 

2.  Write  the  equations  of  the  planes  determined  by  the  following 
data : 

(a)  a  =  3,  6  =  4,  c  =  —  ^  ; 

(6)  p  =  5,  cos  a  =  4,  cos  ^  =  —  ^; 

(c)  through  the  points  (1,  1,  1),  (2,  -  1,  -  3),  (4,  3,  -  2)  ; 

{d)  parallel  to  the  plane  3x— 4?/  +  8^—  5  =  0  and  containing  the 
point  (2,  1,  -  4)  ; 

(e)  perpendicular  to  the  Diane  3.r  —  4?/  +82  —  5  =  0  and  contain- 
ing the  points  (2,  1,  —  4)  and  the  origin  ; 

(/)  having  the  foot  of  the  normal  axis  at  the  point  (—2,  —  4,  1)  ; 

(g)  parallel  to  the  ^r-ayis  and  containing  the  points  (2,  —  4,  1)  and 
(3,5,  -2). 

3.  Find  the  angle  between  each  of  the  planes  of  Problem  1  and  the 
plane  8x  +  2?/-f22-5  =  0. 

4.  Find  the  angle?  at  which  each  of  the  planes  of  Problem  1  meets 
the  coordinate  planes. 

*^5.    Show  that  the  distance  from  the  plane  x  cos  «+  y  cos  ^  -^  z  cos  y/TTi  5 

—  j9  =  0  to  the  point  Cxi,  yi,  zi)  is  Xi  cos  «  +  2/1  cos  ^  -}-  zi  cos  y  —  p. 

^  6.    Using  the  result  of  Problem  5  find  the  distance  from  each  of  the 
planes  of  Problem  1  to  the  point  (3,  0,  —  2). 

7.    Find  the   locus  of  points  equidistant    from    the    points   (—2 
-3,  1)  and  (4,  5,   ^  7). 


192  ANALYTIC   GEOMETRY 

\^e^  '    ^'    ^^^^  *^®  point  of  intersection  of  the  planes 

2x  -Sy  +  2z  =  5; 
Sx  +  by-3z  =  6; 
4:X  —  2y-\-z  =  9. 

9.  Show  that  the  planes  bisecting  and  perpendicular  to  the  edges 
of  the  tetrahedron  whose  vertices  are  (3,  1,-2),  (4,  2,  —  1),  (6,  1,  —5), 
(1,  3,  2)  meet  in  a  point. 

10.  Find  the  locus  of  points  equidistant  from  the  planes  2x  —  By 
+  0=4  and  Sx  —  2y-\-s  =  6. 

11.  Find  the  volume  of  the  tetrahedron  whose  vertices  are  given  in 
Problem  9. 

i    12.    What  is  the  locus  of  each  of  the  following  equations  : 
/  (a)  x2-3x  +  2  .-=0; 

.  (b)x^-y^  =  0', 

(c)  x^  -\-  2  xy  +  7j^  -  z^  =  0? 

ilZ.    Name  two  of  the  following  planes  which  are  (a)  parallel;  (5) 
perpendicular  to  each  other  : 

2x-Sy  +     bz  =  S; 

4:X  +  Qy  +     22=1; 

6  a:  —  9  V  +  15  0  =  5. 

158.  The  Equations  of  the  Line.  —  Since  any  straight  line 
may  be  regarded  as  the  intersection  of  two  planes,  it  will 
be  seen  from  §§  152  and  loo  that  it  requires  two  equations 
of  the  form  Ax+By  +  Cz-hD  =  0 

to  determine  a  straight  line. 

It  is  more  convenient,  ho^^^ver,  to  determine  a  line  from 
two  given  points  upon  it,  or  from  one  point  and  its  direction 
angles.  From  these  we  derive  equations  for  the  line.  It 
should  be  noted  that  in  each  case  two  equations  are  required. 

159.  The  Symmetrical  Equations.  —  Let  the  line  pass 
through  the  point  Piix^,  ih,  z^  and  have  direction  angles 
a,  3,  and  y.  If  P(x,  y,  z)  is  any  point  on  the  line  and  d 
denotes  the  distance  PiP,  we  have  from  (58) 

X  —  X,            n      V  —  Vi                   z  —  z, 
cos  a  = ^,  cos  y8  =  -'       -^S  cos  y  = i- 


d      '         "^  d     '         '  d 


SOLID   ANALYTIC   GEOMETRY  193 


These  relations  give 


cos  a        cos  p       cos  *Y 

which  are  the  symmetrical  equations. 

If  a  point  Pi  and  the  direction  numbers  of  the  line  a,  h, 
and  c  are  knowTi,  a  more  convenient  form  to  use  is 

(67a) 


160.  The  Two-point  Equations.  —  Let  the  line  be  deter- 
mined by  the  points  Pi(a?i,  y^,  z^)  and  F^ix^  yo  z^.  Then 
from  (58)  the  difference  of  the  respective  coordinates  are 
direction  numbers.  Hence  the  two-point  equations  may 
be  obtained  from  (67a)  by  substituting  x^  —  x^,  2/1  —  2/2>  and 
2^  _  z^,  for  a,  b,  and  c,  respectively.     They  are  ^  . 

x  —  x^  _  y  —  y^  __  z  —  z^  (676) 

^  '  X1-X2         1/1-1/2         ^l--^2 

161.  The  Projection  Forms.  —  Let  the  equations  of  the 
projections  of  the  given  line  on  the  xy-  and  ct'2-planes  respec- 
tively be  .      , 

^  y  =  kx-{-m,  ^Q^^ 

z  =  Ix  -\-  n. 

These  equations  taken  together  determine  the  line. 

For  each  equation  may  be  regarded  as  the  equation  of 
the  projecting  plane.  For  example,  the  first  by  (61a)  is  the 
equation  of  a  plane  parallel  to  the  2-axis  and  hence  perpen- 
dicular to  the  a;2/-plane.  But  it  contains  all  points  satisfy- 
ing the  relation  y  =  kx  -\- m  and  so  must  contain  the  given 
projection.  Similarly,  the  other  equation  is  that  of  the 
plane  projecting  the  line  on  the  a;;2-plane.  Thus  the  line  is 
determined  as  the  intersection  of  its  projecting  planes. 


194  ANALYTIC    GEOMETRY 

162.    Reduction  of  the  General  Equations  of  a  Line  to  the 
Standard    Forms.  —  Consider    the    line    determined    by   the 

equations 

4a; -h?/- 2  + 2  =  0, 

ic  +  4  2/  +  22  —  1  =  0. 

To  reduce  these  to  the  projection  forms,  eliminate  first 
z  and  then  y  between  the  equations.     We  have  at  once, 

3x'  +  2^  +  l=0,     5a;-22  +  3  =  0. 

That  the  locus  of  these  equations  is  the  same  as  the  locus  of 
those  given  is  obvious,  since  coordinates  satisfying  the  first 
pair  of  equations  must  satisfy  the  last. 

To  reduce  the  given  equations  to  the  symmetrical  form 
solve  the  projection  forms  for  x.     We  have, 

2^+1  2  2  -  3    *< 

Equating  these,  we  have 

1—3  5       * 

+  22 

The  denominators  are  direction  numbers  ;  and  if  we  divide 

each  one  by  Vl^  +(—  f)^  +  {W  =  '^^^^  ^^^J  become  direc- 
tion cosines  by  §  150.     Doing  so,  we  get 


-3 

V38       V38       V38 

which  are  the  symmetrical  equations  of  a  line  having  direc- 
tion cosines   -7=,  -==:,  -7=.   and  passing  through  the    point 

V3S     V3S      V3S 

(0,  —  i,  4).  But  these  are  merely  transformations  of  the 
original  aquations,  and  so  are  the  symmetrical  forms  re- 
quired. 


SOLID   AXALYTIC   GEOMETRY  195 


PROBLEMS 

1.  Find  the  points  at  which  the  following  lines  cut  the  coordinate 
planes  and  draw  the  lines  : 

(a)   2x-Sy  +  2z-6  =  0,Sx-y-z  +  e=0; 

(6)    2x-4?/  =  5,  Sy-z  =  2; 

,  .   y-2     ?/  -  3     g+  1 

(c)   =  - = • 

^  ^       6  2  3 

2.  Eeduce  the  equations  in  Problem  1  to  the  symmetrical  form. 

3.  Reduce  the  equations  in  Problem  1  to  the  projection  form,  the 
projecting  planes  being  perpendicular  to  the  xz-  and  yz--p\&nes. 

4.  Find  the  equations  of  the  straight  lines  determined  as  follows  : 
(a)  through  the  points  (1,  0,  —  5)  and  (—  2,  3,  1)  ; 

(6)  through  the  point  (1,  —  1,  2)  and  parallel  to  the  s-axis; 

(c)  through  the  point  (1,  —  1,  2)  and  perpendicular  to  the  2- axis  ; 

(d)  through  the  point  (1,  -  1,  2)  and  having  cos  a  =  i,  cos  /3  =  \. 

(e)  through  the  point  (1,  —  1,  2)  and  parallel  to  the  line  of  Prob- 
lem 1  (a). 

5.  Do  the  lines  of  Problem  1  (a)  and  (6)  meet  ? 

6.  Show  that  if  a  line  is  perpendicular  to  a  plane,  it  has  the  same 
direction  cosines  as  the  normal  to  the  plane. 

7.  Show  that  if  a  line  is  parallel  to  a  plane  its  direction  cosmes 
and  those  of  the  normal  to  the  plane  satisfy  the  relation 

cos  a  cos  a'  +  cos  /3  cos  ,3'  +  cos  y  cos  7'  =  0. 

8.  Find  the  angle  between  the  lines 

x-S^y-\-2^z-o  ^^^  o-  +  S^y-2^z-4: 
2  3  6  6  3  1 

9.  Show  that  the  line  ^^^  =  ^^  =  ^^ 

(a)  is  parallel  to  the  plane  6x  —  Sy  -{-  12  z  -  60  =  0  ; 

(6)  lies  in  the  plane  Sx-iy-\-Qz-19  =  0; 

(c)  is  perpendicular  to  the  plane  2  x  -\- ?>  y  -{-  z  =  6. 

10.  Find  in  the  symmetrical  form  the  equations  of  the  locus  of 
points 

(a)  equidistant  from  the  points  (3,  -1,  2),  (4,  —6,  -5),  and 
(0,  0,  -  3)  ; 

(6)  equidistant  from  the  planes  2x  -3y=6,  Sx  -2z  =  8,  and 
2x  +  32=:5. 


196 


ANALYTIC   GEOMETRY 


163.  Cylindrical  Surfaces.  —  A  cylindrical  surface  is  a  sur- 
face generated  by  a  moving  straight  line  which  constantly  in- 
tersects a  given  fixed  curve  and  remains  parallel  to  a  fixed 
stnttght  line.  The  fixed  curve  is  called  the  directrix  and  the 
generating  line  the  generatrix. 

Let  the  directrix  of  a  cylindrical  surface  be  an  ellipse  in 
the  ir?/-plane  sj^mmetrical  with  respect  to  the  coordinate  axes, 
and  let  the  generatrix  remain  parallel  to  the  z-Sixis.     Then 

the  equation  of  the  ellipse 
in  its  plane  will  be 

b'^x-  +  aY  =  a^W. 

Let  P{x^  y,  z)  be  any 
point  on  the  surface,  and  Q 
the  corresponding  point  on 
the  ellipse.  Evidently  the 
coordinates  of  Q  will  be 
{x,  y,  0)  ;  i.e.  the  x-  and  y- 
coordinates  of  every  point 
on  the  line  PQ  are  the  same 
as  those  of  Q.  But  the  co- 
ordinates of  Q  satisfy  the 
equation  of  the  ellipse.  Hence  for  any  point  on  the  cylin- 
drical surface,  52^^,2  _^  ,^2^2  ^  ^252. 

The  preceding  is  perfectly  general,  and  we  see  that  it 
leads  at  once  to  the  theorem  : 

TJie  locus  of  a  cylindrical  surface  ivhich  has  for  its  directrix 
a  curve  in  the  xy-plane  and  ivhose  generatrix  moves  parallel 
to  the  z-axis,  lias  the  same  equation  as  the  directrix. 

The  converse,  which  is  readily  established,  is  i 
Hie  locus  of  an  equation  in  two  variables,  x  and  y,  is  a 
cylindrical  surface,  ivhose  directrix  is  the  curve  in  the  xy-plane 
which  has  the   same   equation,  and   whose  generatrix  moves 
parallel  to  the  z-axis. 


SOLID   ANALYTIC   GEOMETRY 


197 


Similar  theorems  hold  of  course  for  cylinders  whose  gen- 
eratrices are  parallel  to  either  the  oc-  or  ^/-axes. 

Exercise.     What  is  the  locus  of  any  equation  in  one  variable  ? 

164.  Surfaces  of  Revolution.  —  A  surface  generated  by  re- 
volvinq  a  plane  curve  about  a  fixed  line  in  its  plane  as  an  axis 
is  called  a  surface  of  revolution. 

The  curve  is  called  the  generatrix ;  its  position  with  ref- 
erence to  the  axis  is  unchanged  during  the  revolution. 
Sections  of  the  surface  made  by  planes  through  the  axis  are 
called  meridional  sections.  Erom  the  definition  of  such  a 
surface  it  is  evident  that 

(a)  sections  made  by  planes  perpendicular  to  the  axis 
are  circles ; 

(h)  any  meridional  section  is  the  generatrix  itself. 

Let  us  first  consider  a  conical  surface  generated  loy  re- 
volving a  straight  line  about  the  2;-axis.  Let  the  position 
of  the  generatrix  in  the  x2;-plane  be  AB,  of  which  the 
equation  is 

2  a;  +  2  =  5. 

Let  P{x,  y,  z)  be  any  point 
of  the  locus.  As  the  line 
turns  about  the  2;-axis,  P 
describes  a  circle  of  radius 
r  =  LP  in  a  plane  parallel 
to  the  a'2/-plane  and  distant 
from  it  KP  =  z.  Now  when 
the  line  is  in  the  x2:-plane, 
x=r',  hence  for  all  posi- 
tions of  the  line, 

2r  +  z  =  5. 

But,  as  P  describes  a  circle  of  radius  r  in  a  plane  parallel 
to  the  xy'-plsine,  we  have  at  once 


b<^ 


x^  +  y 


2  ^ 


^ 


198 


ANALYTIC    GEOMETRY 


Substituting  above,  we  have 

2  V.^c^TF  -{-z  =  5. 

Simplifying,  we  have  the  equation 

4(a;2+  7/)  =  (z-5y. 

Consider  the  surface  generated   by  revolving   about  the 
ic-axis  a  circle  of  radius  a  whose  center  is  the  origin.     The 

equation  of  the  generatrix 
in  the  a.'2;-plane  is 

x2  +z''  =  a\ 

Evidently  the  point  P 
describes  a  circle  with 
its  center  on  the  a>axis  and 
of  radius  r ;    hence 

X^  +  7'2  _  ^j2^ 

But  2/-  +  2^  =  r\ 

Hence  we  have  at  once, 

x2  +  2/2  +  22  =  a\     (69) 

This  equation  is    important,  as  it  is  the  equation  of  a 
sphere  of  radius  a  and  having  its  center  at  the  origin. 


PROBLEMS 

1.  Identify  and  sketch  the  following  surfaces  : 
(a)  a:2-4^2  =  4;  {d)  xz  =  15  ; 

(6)  a:2  +  2  s2  =  2  ;  (e)  ^2  _,_  4  ^  _  5  ^  0  ; 

(c)  y2^z2_^2y  +  62;  =  15  ;  (/)  x^  =  16 z. 

2.  Find  the  equation  of  the  surface  generated  by  revolving  the 
curve  about  the  axis  indicated  : 

(a)  Ho'  -{-12y  =  3,  cc-axis 

(6)  z^  =  2px,  a:-axis 

(c)  h-^x'^  +  a~z-  =  a'^b^,  a:-axis 

(d)  b-x^  +  a-z^  =  a'^b-\  z-axis 

(e)  b^oo^  —arz^  =  a^b"^,  a;-axis 

(f)  b-^x'^  -  a^z^=a^b%  ^-axis 


SOLID   ANALYTIC   GEOMETRY  199 

3.  Find  the  equation  of  the  surface  of  a  cone  whose  vertex  is  at  the 
origin,  with  the  0-axis  for  its  axis  of  revolution  and  opposite  elements 
perpendicular  to  each  other. 

4.  Find  the  equation  of  the  locus  of  a  point  equidistant  from  a 
given  plane  and  a  given  line  parallel  to  the  plane. 

5.  Show  that  the  locus  of  a  point  equidistant  from  the  point 
(p,  0,  0)  and  the  ?/2-plane  is  the  paraboloid  of  revolution  y-  +  z^ 
=  '2px  —p^. 

6.  Show  that  the  locus  of  a  point  in  space  the  sum  of  whose 
distances  from   the  points    ( ±  c,  0,  0)    is  2  a  is  the  prolate  spheroid 

a^     62      52 

r-       )/■-       z- 

7.  How  mav  the  hyperboloid  —  _^  —  —  =  1  be  defined  as  a  locus  ? 

a-      6-     C- 


165.  Discussion  of  Surfaces.  —  The  discussion  of  a  sur- 
face is  ill  general  a  much  more  complicated  matter  than  that 
of  a  curve.  The  notions  of  intercepts,  symmetry,  and  ex- 
tent are  readily  generalized.  Besides  these,  however,  we 
have  two  other  notions,  which  are  of  help  in  determining 
the  nature  of  a  surface.  The  first  is  the  section  of  the  sur- 
face by  any  plane  parallel  to  one  of  the  coordinate  planes  ; 
the  second  is  the  section  of  the  surface  by  any  of  the  coor- 
dinate planes.     The  latter  is  called  a  trace  of  the  surface. 

To  illustrate  these,  consider  the  sphere,  with  equation 


x"  +  2/-  +  ^-  =  a\ 

To  find  the  equations  of  the  traces,  set  x,  y,  and  z  succes- 
sively equal  to  0.     When  x  =  0,  we  have 

2/2  +  ^2  =  a\ 

Thus  the  2/^-trace  is  a  circle  of  radius  a.     The  xy-  and  xz- 
traces  are  also  circles  of  the  same  radius,  with  equations 

.T^  -f-  ?/2  =  a^ 
and  Qt?  -f  z-  =  a-  respectively. 


200  ANALYTIC   GEOMETRY 

The  section  made  by  a  plane  parallel  to  the  ?/2;-plane  and 
at  a  distance  c  from  it  is  the  curve  determined  by  the  given 
equation  and  the  equation  x  =  c.  Eliminating  x  between 
the  two  equations,  we  have 

?/2  -}-  2;2  =  a^  —  c^, 

which  shows  that  the  section  is  a  circle  having  the  point 
(c,  0,  0)  as  a  center  and  Va^  —  c^  as  the  radius.  If  c'>  a, 
this  is  imaginary,  indicating  that  the  surface  lies  wholly 
within  the  planes  x  =  ±  a.  A  similar  discussion  holds  for 
sections  parallel  to  the  other  coordinate  planes. 

To  sketch  a  surface  when  only  two  variables  of  its  equa- 
tion are  of  the  same  degree,  or  of  the  same  degree  and  sign, 
first  draw  sections  parallel  to  the  plane  of  the  two  variables. 
These  sections  and  the  traces  give  the  best  representation 
of  the  locus. 

166.  Quadric  Surfaces.  —  By  analogy  to  the  conies,  the 
locus  of  the  general  equation  of  the  second  degree 

Ax^  +  By^  +  Cz^  +  Dxy  +  Eyz  +  Fxz+Gx  -\-Hy+Kz+L  =  0 

is  called  a  quadric  surface.  It  can  be  shown  that  these  sur- 
faces have  properties  analogous  to  those  of  the  conies  ;  and  in 
particular  that  every  plane  section  of  such  a  surface  is  a 
conic.  We  shall  not  go  into  these  details,  but  will  confine 
our  discussion  to  the  simpler  forms  to  which  the  general 
equation  may  be  reduced. 


167.   The  Ellipsoid.  —  The  locus  of  the  equation 

x2       y2       22 

— h  —  H — 
a2      bi      c2 


r2       1/2        72 

-,  +  r-,  +  T,  =  l  ('0) 


is  called  an  ellipsoid. 

From  the  form  of  the  equation  we  see  at  once  that  the 


SOLID   ANALYTIC   GEOMETRY 


201 


surface  is  symmetrical  with  respect  to  all  of  the  coordinate 
planes  and  coordinate  axes.     The  intercepts  on  the  axes  are 

X  =  ±a, 

2/  =  ±  ^ 

z  =  ±  c. 

Setting  x  =  0,  the  equa- 
tion of  the  trace  on  the 
?/2-plane  is 

hence  it  is  an  ellipse  with 

semi-axes    b    and    c.      In 

like    manner   the   traces   on   the   other   coordinate   planes 

are  ellipses. 

Settinsf  x  =  k,  the  equation  becomes  ^-f  —  = — -,   a 

form  which  shows  that  there  is  no  section  for  k  numerically 
greater  than  a.  Dividing  by  the  right-hand  member,  this 
becomes 


r 


=  1. 


Hence  the  various  sections  are  ellipses  symmetrical  with 
respect    to    the    ic-axis.     The    semi-axes,    -  Va^  _  k"^    and 


a 


£.  V<^2_;^2^  ^^Q^  smaller  as  k  increases,  until  the  ellipse  be- 
et 

comes  a  point  when  k  =  a.     Similarly  the  sections  parallel 
to  the  other  coordinate  planes  are  ellipses. 

The  values  a,  b,  and  c  are  called  the  semi-axes  of  the 
ellipsoid.  When  a  =  5  =  c,  the  ellipsoid  is  evidently  a  sphere 
with  the  center  at  the  origin.  Ordinarily  the  semi-axes  are 
unequal,  and  in  this  form  of  the  equation  it  is  usually  as- 
sumed that   a  >  6  > c.     When   6  =  c,   but   a>b   or  c,  the 


202 


ANALYTIC   GEOMETRY 


ellipsoid  is  called  a  prolate  spheroid.  In  this  case  the  sec- 
tions parallel  to  the  2/2;-plane  are  circles.  The  locus  is  then 
a  surface  of  revolution  since  it  is  generated  by  revolving 
the  a;2;-trace  (or  the  x7j-tTSice)  about  the  a:-axis.  When  a  =  h 
but  c<Ca  or  6,  the  surface  is  that  of  an  oblate  spheroid. 
This  is  also  a  surface  of  revolution  and  the  sections  parallel 
to  the  a?^-plane  are  circles. 

168.   Hyperboloid  of  One  Sheet.  —  The  locus  of  the  equation 

^+^-"^=1  (71) 

is  called  a  hyperboloid  of  one  sheet. 


^ 

- 

Z 

A      ^ 

V 

/ 

-V 

\ 

/ 

1 

i                  N 

/ 

J 

1       '-'r'    / 

/ 

A'K 

'%> 

X 

/ 

-|-V 

1  / 
1  / 

-\ 

V. 



•■    - 

^ 

In  this  case  sections  parallel  to  the  a;?/-plane,  made  by 
the  planes  2  =  A:,  have  equations 


y 


W- 


a^      IP-  (^ 

and  hence  are  ellipses  which  increase  in  size  as  the  numerical 
value  of  k  increases. 


SOLID   ANALYTIC   GEOMETRY 


203 


The  cc^-trace  and  the  yz-tmce  are  the  hyperbolas 

/>'2  y2  9/2         J.2 

__^  =  1  and  •'^  —  -  =  1  respectively. 
a"      e  fr      & 

If  a  =  6,  the  locus  is  a  surface  of  revolution  about  the 

2;-axis. 


The    locus    of    the 

my 


169.   Hyperboloid    of    Two    Sheets. 

equation  x^     y-      ^'  ^-t 

is  called  the  hyperboloid  of  two  sheets. 

In  this  case  we  consider  sections  parallel  to  the  2/2-plane 
made  by  the  planes  x  =  k.     These  are  the  ellipses 


If  k^  <  o?  the  right-hand  member  is  negative.  Hence  no 
part  of  the  locus  is  between  the  two  planes  x  =  ±  a.  If 
k  =  ±a,  each  corresponding  section  is  a  point  ellipse.  As 
k  increases  numerically  from  a  to  go  the  section  increases 
indefinitely. 

The  .T?/-trace  and  the  ajg-trace  are  the  hyperbolas 

^  _  ^  =  1  and  -  —  -  =  1  respectively, 
a-     52  ^2     c2 

If  6  =  c,  the  locus  is  a  surface  of  revolution. 


204 


ANALYTIC    GEOMETRY 


170.  Elliptic  Paraboloid.  —  This  is  the  locus  of  the  equa- 
tion       ^       ^  :>-tAr^^ 

-  +  ^'  =  2  cz.       (73) 

Proceeding  as  before,  we 
observe  that  sections  made 
by  the  planes  z  =  'k  are  el- 
lipses whose  axes  increase 
indefinitely  as  A;  increases. 
The  surface  lies  wholly 
above  or  wholly  below  the 
a'2/-plane,  according  as  c  is 
positive  or  negative.  The 
xz-  and  2/2^-traces  are  parab- 
olas. If  a  =  5,  the  locus 
is  a  surface  of  revolution. 

171.  Hyperbolic   Paraboloid.  —  This    is    the   locus  of  the 

equation  2      ,,9 

^-^  =  2cz.  (74) 


Consider   c  positive.     Sections  made  by  the  planes  2;  =  A: 
are  the  hyperbolas 

As  Tc  increases  from 
0  to  00  the  vertices  of 
the  corresponding  sec- 
tions lie  in  the  .T2:-plane 
and  recede  indefinitely 
from  the  2-axis.  As  k 
decreases  from  0  to  —  00 
the  vertices  of  the  sec- 
tions are  in  the  ?/2;-plane  and  recede  indefinitely  from  the 
z-axis.     The  xz-  and  ?/2-traces  are  parabolas,  each  having  its 


SOLID   ANALYTIC   GEOMETRY  205 

vertex  at  the  origin,  the  former  extending  above  the 
a-^z-plane,  the  latter  below.  The  xy-tmee  is  a  pair  of  lines 
intersecting  at  the  origin. 

PROBLEMS 

1.  Find  the  traces  of  the  following  surfaces  on  each  coordinate 
plane : 

(a)  y^-\-z^  =  l2z;  (e)  x^  -  4:y^  -  Az^  =  4: ; 

(6)  x^-y^  =  0;  (/)  «-  +  y-=Qz-; 

(c)  x2  +  4?/2  +  4^2  =  4  ;  (g)  x^  +  4y^ -{-2z^  =  W  ; 

id)  x^  +  42/-^  -  4^2  =  4  ;  (h)  x^  =  y^  +  9 z\ 

2.  Find  the  intersection  of  each  of  the  above  surfaces  and  the 
plane  .r  =  4. 

3.  Discuss  the  traces  of  the  following  surfaces  : 

a-      0- 

4.  Show  that  the  sections  of  —-{-^-~=z  are  parabolas  if  perpendic- 

a^     0- 

ular  to  the  x-  or  the  ?/-axis,  and  ellipses  if  perpendicular  to  the  ^-axis 
and  5  >  0. 

5.  Show  that  the  sections  of  the  surface  ^-  ^  =  2  perpendicular 

to  the  X-  or  the  ?/-axis  are  parabolas.  How  do  those  perpendicular  to 
the  :f-axis  differ  from  those  perpendicular  to  the  ?/-axis  ? 

6.  Discuss  and  sketch  the  following  surfaces  : 

(a)  4a:-^+25?/2  +  16  2-^rr  100; 
(6)  28  x2  +  196  =  9  ?/-2  +  16  z^  ; 
(c)  4.r2  +  9!/2_^2  +  36  =  0; 
(fO  16  .r2  +  ?/2  ^  04  ^  33  0  ; 
(e)  100^2 +  .36?/2-81x  =  0; 
(/)   ?/2_  22  +  9a;  =  0. 


206  ANALYTIC   GEOMETRY 

FORMULAS   AND    EQUATIONS 
Formulas  of  Distaxce  and  Direction 

Xo.      Page 

Distance:  d  =  -^{x^- x^f  +  iy^-y^f         (1)       12 

Point  dividing  line  in  the  ratio  i\  :  r^ : 

r^x^±r^^  y^^nill±Ml.  (2)       13 

ri  4-  r2  Vi  H-  rg 

Mid-point :         a^^  =  J  (x^  +  a-g),  ?/o  =  2  (2/1  +  2/2).         (2  a)       13 
Slope:  m=-^/-y^  (^)       ^' 

X-[ iCo 

Test  for  parallelism  :  mi  =  m2.  (4)       18 

Test  for  perpendicularity :        mi7?i2  =  —  1.  (5)       18 

Angle  beween  two  lines  :        tan  /?  =    '^  ~ — ^.  (6)       18 

1  +  mim.2 

The  Straight  Lixe 
Equations : 

Point  slope  form :  y  ^ y^=  m(x  —  .^2).        (7)       40 

Two  point  form  :  ^ — ^  =  "^^  ~'^^  •  (7  a)      ■  41 

X  —  X^        X^  —  X2 

Slope  intercept  form  :  y  =  7nx  +  b.  (8)       41 

Intercept  form :  ^  +  -^  =  L  (9)       41 

a      b 

General  form :         Ax  +  By-\-C  =  0.  (10)  44 

Test  for  parallelism :  A.A'  =  B:B'.  45 

Test  for  perpendicularity  :  AA'  =  —  BE'.  46 

Test  for  identity :  A :  .4'  =  B:B'=C:C'.  46 

Normal  form :  x  cos  <ii-\-y  sin  o)  —p  =  0.        (11)  51 

To  reduce  form  (10)  to  form  (11),  divide  by 

±  V32"+^  sign  to  agree  with  that  of  B.  52 


FORMULAS  AND   EQUATIONSk,^^     207 


Distance  from  a  line  to  a  point :        M  '    -^  ^"^V     no.     Page 
d  =  Xi  cos  CO  H-  ?/i  sin  w  —p.  (12)       54 


The  Circle 

Equations  : 

Standard  form  :        {x  -  lif-^  {y  -  ky  =  r\ 
General  form  :      x^ -\-y^  +  Dx  +  Ey  +  F=  0. 

Length  of  a  tangent  from  Pi : 

t2  =  x^^-^yi'  +  Dx,  +  Ey,-\-  F. 

Equation  of  the  radical  axis  : 

(D-D'yx-\-{E-E')y-\-F-F'  =  0. 


(13)  64 

(14)  ■  66 

(15)  70 
(15  a)  71 

(16)  74 


The  Parabola 

Equations : 

2/2  =  2px. 

(17) 

78 

a;2  =  2  py. 

(17  «) 

78 

Latus  rectum : 

2p 

79 

The  Ellipse 

Formulas  connecting  the  fundamental  constants  : 

a 
c-\-p=-' 
e 

c=  ae. 

a2  =  62  _^  c2. 

,2 


Equations : 

Focal  radii : 
Latus  rectum : 


+  'f=l- 


62 


x"- 
a2 

f.o^^  1 
a2      62 

p-\-p'  =  2a. 

o  2  62 

2  ep  = • 

a 


(18) 

85 

(19) 

85 

(21) 

87 

(20) 

86 

20  a) 

90 

89 

90 

208  ANALYTIC  GEOMETRY 

The  Hyperbola 
Formulas  connecting  the  fundamental  constants  : 


Equations : 


No. 

Page 

a 

C-p=-' 

e 

(22) 

96 

c  =  ae. 

(23) 

96 

c^  =  a^ 

+  62. 

(25) 

98 

»2        7/2  _ 

a2     62~    • 

(24) 

98 

Latus  rectum :                   2  ej?  =  *"     • 

a 

99 

Focal  radii :                   p  —  p'  =  2a. 

101 

Equations  of  the  asymptotes  : 

a 

(26)     102 

x^      if  _  ^ 
a2      62 

(26  a)     102 

Traxsformatiox  of  Coordixates 

Translation  :  x  =  x'  +  h,  y  =  1/  -\-k.  (27)     108 

Rotation  :  x  =  x'  cos  0  —  y'  sin  0, 

y  =  x'  sin  6  -\-y'  cos  ^.  ^"^  -^ 

Equation  of  the  conic  with  the  directrix  as  the 

2/-axis  and  the  focus  on  the  i»-axis  : 

{x-2->y  +  y''=e^x\  (29)     110 

Generalized  standard  equations  of  the  conies  : 

'^ihirabola:  {y -kf  =  2 p{x-h).  (30)     117 

{x-liy  =  2p{y-k),  (30a)     117 


FORMULAS  AND   EQUATIONS  209 

No.        Page 

Ellipse:                (^sd!Y  +  (jL=^  =  l,  (31)     118 

(y-ky_^(x-hy^^^  (31a)     ng 

Hyperbola:         (^^-(^^  =  1.  (32)     118 

Angle  of  rotation  for  eliminating  tlie  xy  term : 

tan  2^  =  -^.  (33)     120 
A  —  G 


Polar  Coordinates 

Eelation  to  rectangular  coordinates  : 

x  =  pcos^,  7/  =  psin^.  (34)     129 

^2  =  0^2+2/2,  ^  =  arc  tan '^.  (35)     129 

Equations  of  the  straight  line : 

pcos^  =  a.  (36)  131 

p  sin  ^  =  a.  (36  a)  131 

e=c.  (37)  131 

Equations  of  the  circle  : 
^  p=r.  (38)     131 

p  =  2rcos^.  (39)  131 

p  =  2r  sin  6.  (39  a)  131 

p  =  acos^  +  6sin(9.  (40)  132 

Formula  of  rotation :         0  =  6'  + «.  (41)  136 

Equations  of  the  conies  : 

p^^lL^.  (42)     138 

1  —  e  cos  6 

p  = ^-E (42  a)     138 

1  —  e  sin  6 


210  ANALYTIC   GEOMETRY 

Tangents  and  Xoemals 
Slope  of  a  tangent :  No.      Page 

To  the  circle :  m  =  -^-  (43)     167 

yi 

To  the  ellipse :  m  =  -—''  (44)     168 

To  the  parabola :  m  =  i--  (45)     168 

2/1 

To  the  hyperbola :         m  =  ^.  (46)     168 

Length  of  the  subtangent :         ^-  (47)     171 

m 

Length  of  the  subnormal :         myi.  (48)     171 

Equation  of  the  diameter  bisecting  chords  of  slope  m : 

In  the  ellipse  :  y  = x.  (49)     177 


In  the  parabola :  yz=i^.  (51)     177 

m 

In  the  hyperbola :  y  = x.  (52)     177 

ahn 

Relation  between  the  slopes  of  conjugate  diameters  : 

?  - 
In  the  ellipse :  nwi'  = (50)     177 

In  the  hyperbola :      mm'  =  —'  (53)     178 

Solid  Analytic  Geometry 

Radius  vector  and  direction  cosines : 

p  =  a.'2  +  2/2_^22,  (54)  182 

x  =  p  cos  a,  y  =  p  cos  ft,  z  =  p  cos  y.  (55)  183 

cos'^  a  +  cos^  /?  +  cos^  7  =  1-  C^^)  *■  ^^^ 


FORMULAS  AND   EQUATIONS  211 

Distance :  No      page 

d  =  VCx-i  -  x,y  +(2/i  -  y,y  +  (^1  -  z,)\      (57)^183 
Direction  cosines  : 

cos«  =  ?i.=^=,  cos  ;8=?-^!-=^S  cosy  =  ?L=L^.     (58)^184 
d  d  cl 

Angle  between  two  lines  : 

cos  ^=cos  «!  cos  cco+cos  ^1  cos  /32+cos  yi  cos  y2-     (59)  //185 
Condition  for  parallelism : 

«i  =  «2j  P\  =  ^-2^y\  =  7-2,  or 

«^  =  TT  —  tto,    /?!  =  TT  —  /?.2J    yi  =  -n-  —  72- 

Condition  for  perpendicularity : 

cos  «i  cos  «2  +  cos  ^1  cos  ^2  +  cos  yi  COS  yo  =  0.   (60)  *^85 

Equations  of  the  plane  : 

Normal  form  :     a;  cos  «  +  ?/ cos /?  +  2;  cos  y  =  p.     (61)  »^188 
General  form  :  Ax  +  By  +  Cz  +  D  =  0.     (62)  ^lB9 

Angle  between  two  planes  : 

Test  for  parallelism : 

A:A'  =  B:B'=C:C'.  (64)  <L90 

Test  for  perpendicularity : 

AA'  +  BB'+CC  =  0.  {65)  ^^90 

Intercept  equation  of  the  plane  : 

yi^   -  V  e:  +  ^  +  ?=l.  ■    (66)/l91 

a     b     c 

Equations  of  the  line : 

Symmetrical  forms  :  '^^=^  =  ^^  =  ^^^  •        (67)^-^3 

cos  a       cos  (3       cos  y 

£zi£i  =  t:li  =  ^ziii.     (67a)Kf93 
a  b  c 


212  ANALYTIC   GEOMETRY 

No.         Page 

Two  point  forms  :      ^  -  ^i  =  U^zIl  =  IZLh. .    (67  h)  »493 

^1-^*2  2/1-^2         ^1-2^2 

Projection  forms  :         y  =  kx-{-  m,  z=lx-\-  n.       (68)  y^93 
Equations  of  the  quadric  surfaces  : 

Sphere:  x^  +  y^ -\- z""  =  a\        (69)     198 

Ellipsoid:  ^^^t^t  =  i,         (70)     200 


a2      b 


(. 


2  2  9 

Hyperboloid  of  one  sheet :     -  +  -^  —  -  =  1.  (71)     202 

a^      ^"      (,'2 

2  2  2 

Hyperboloid  of  two  sheets  :  —  _  -^  _  ^  =  1.         (72)     203 

a^      If-      c^ 

Elliptic  paraboloid :  ^  +  ?!^"  =  2c;s.     (73)     204 

CL^      IP- 

Hyperbolic  paraboloid :  -  —  ^  =  2  c;^.      (74)     204 


p/3 

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f^3 

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pi-? 

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Tr/uvt 

f  ^S^- 

fa.   . 

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P  7-^' 

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